{"id":24490428,"url":"https://github.com/alishahusain/newt","last_synced_at":"2025-06-13T14:02:31.514Z","repository":{"id":60383526,"uuid":"535170171","full_name":"alishahusain/newt","owner":"alishahusain","description":"๐Ÿธ pattern algorithms data structures in javascript pseudo code space time complexity and most common implementation of each algorithm and data structure 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\n\n## 14 Patterns Algorithim\n\u003ctable\u003e\n\u003cth\u003e\u003ch3\u003eUses: Symbol in Pattern\u003c/h3\u003e\u003c/th\u003e\n\u003ctr\u003e\n\u003ctd\u003e\n\u003ch4\u003e๐Ÿ“Œ = 1 Pointers \u003ch4\u003e \n\u003c/td\u003e\n\u003ctd\u003e\n\u003ch4\u003e๐Ÿ“ฆ = 1 Heap \u003ch4\u003e \n\u003c/td\u003e\n\n\u003ctd\u003e\n\u003ch4\u003e๐Ÿšฅ = 1 Array \u003ch4\u003e \n\u003c/td\u003e\n\n\u003ctd\u003e\n\u003ch4\u003e๐Ÿฝ = 1 Stack\u003c/h4\u003e\n\u003c/td\u003e\n\n\u003ctd\u003e\n\u003ch4\u003e๐Ÿฆœ = 1 Queue\u003c/h4\u003e\n\u003c/td\u003e\n\n\n\u003ctd\u003e\n\u003ch4\u003e๐Ÿ—‚ = 1 Hash \u003c/h4\u003e\n\u003c/td\u003e\n\n\u003ctd\u003e\n\u003ch4\u003eโž• = 1 Counter \u003c/h4\u003e\n\u003c/td\u003e\n\n\u003c/tr\u003e\n\u003ctr\u003e\n\n\u003ctd\u003e\n\u003ch4\u003e๐ŸŒฑ = 1 Node \u003c/h4\u003e\n\u003c/td\u003e\n\n\u003ctd\u003e\n\u003ch4\u003e๐ŸŽก = 1 Loop \u003c/h4\u003e\n\u003c/td\u003e\n\n\u003ctd\u003e\n\u003ch4\u003e๐Ÿ“ฑ = 1 Graph\u003c/h4\u003e\n\u003c/td\u003e\n\n\u003ctd\u003e\n\u003ch4\u003e๐Ÿ”— = 1 Linked List \u003c/h4\u003e\n\u003c/td\u003e\n\u003ctd\u003e\n\u003ch4\u003e ๐Ÿชž = 1 Recursion \u003c/h4\u003e\n\u003c/td\u003e\n\n\u003ctd\u003e\n\u003ch4\u003e๐Ÿคนโ€โ™€๏ธ = 1 Swap \u003c/h4\u003e\n\u003c/td\u003e\n\u003c/tr\u003e\n\u003c/table\u003e\n\n \u003cdetails\u003e\n \u003csummary\u003e \u003ch1\u003e๐ŸชŸ Sliding Window : ๐Ÿ“Œ๐Ÿšฅ๐ŸŽก \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n \u003ctable\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003eโ“ SUBSTRING OR SUBARRAY FIND LONGEST OR SMALLEST CONTAIN CHARACTER \u003ch3\u003e\n \u003ch3\u003eโฐ: O(n) ๐Ÿช: O(n) \u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"2\"\u003e\n ๐Ÿฃ Maximum Sum Subarray of Size K, Longest Substring with K Distinct CharactersString Anagrams, No-repeat Substring, etc.\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n\u003cpre class=\"notranslate\"\u003e\n\u003ccode\u003e\nfunction findAvgSubArrays(K, arr) {\n const result = [];\n let windowSum = 0, windowStart = 0;\n for (let winEnd = 0; winEnd \u003c arr.length; winEnd++) {\n windowSum += arr[winEnd]; \n if (winEnd \u003e= K - 1) {\n result.push(windowSum / K); \n windowSum -= arr[windowStart];\n windowStart += 1; \n }\n }\n return result;\n}\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e\n \u003cimg\n src=\"https://miro.medium.com/max/372/0*o-PhzdP_jd_5L2iq.jpg\"\n alt=\"Sliding Window Pattern\"\n /\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n\n\u003c/table\u003e\n\u003cdetails\u003e\n\u003csummary\u003e๐ŸชŸ Problems \u003c/summary\u003e\n\u003cbr\u003e\n## [Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/) #3 ๐ŸชŸ \n โ“: Given a string s, find the length of the longest substring without repeating characters.\n ๐Ÿฃ: 1๏ธโƒฃ Input: s = \"abcabcbb\" Output: 3 Explain: The answer is \"abc\", with the length of 3. 2๏ธโƒฃ Input: s = \"bbbbb\" Output: 1 Explain: The answer is \"b\", with the length of 1. 3๏ธโƒฃ Input: s = \"pwwkew\" Output: 3 Explain: The answer is \"wke\", with the length of 3. Notice that the answer must be a substring, \"pwke\" is a subsequence \u0026 not a substring. #4 Input: s = \"\" Output: 0\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n^3) ๐Ÿช: O(min(m, n)) \n ๐Ÿ‡ Solution: ๐ŸชŸ Sliding Window Pattern โฐ: O(n) ๐Ÿช: O(n)\n\n var lengthOfLongestSubstring = function (s) {\n let result = 0;\n let left = 0;\n let right = 0;\n let set = new Set();\n while (right \u003c s.length) {\n if (!set.has(s[right])) {\n set.add(s[right]);\n right++;\n result = Math.max(result, set.size);\n } else {\n set.delete(s[left]);\n left++;\n }\n }\n return result;\n };\n## [Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/) #424 ๐ŸชŸ \n โ“: You are given a string s \u0026 an integer k. You can choose any character of the string \u0026 change it to any other uppercase English character. You can perform this operation at most k times.\n Return the length of the longest substring containing the same letter you can get after performing the above operations.\n ๐Ÿฃ: 1๏ธโƒฃ Input: s = \"ABAB\", k = 2 Output: 4 Explain: Replace the two 'A's with two 'B's or vice versa. 2๏ธโƒฃ Input: s = \"AABABBA\", k = 1 Output: 4 Explain: Replace the one 'A' in the middle with 'B' \u0026 form \"AABBBBA\". The substring \"BBBB\" has the longest repeating letters, which is 4.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n^2) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐ŸชŸ Sliding Window Pattern โฐ: O(n) ๐Ÿช: O(n)\n\n var characterReplacement = function (s, k) {\n let left = 0;\n let right = 0;\n let maxCount = 0;\n let map = new Map();\n while (right \u003c s.length) {\n map.set(s[right], (map.get(s[right]) || 0) + 1);\n maxCount = Math.max(maxCount, map.get(s[right]));\n if (right - left + 1 - maxCount \u003e k) {\n map.set(s[left], map.get(s[left]) - 1);\n left++;\n }\n right++;\n }\n return right - left;\n };\n## [Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/) #76 ๐ŸชŸ \n โ“: Given two strings s \u0026 t of lengths m \u0026 n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string \"\".\n The testcases will be generated such that the answer is unique.\n A substring is a contiguous sequence of characters within the string.\n ๐Ÿฃ: 1๏ธโƒฃ Input: s = \"ADOBECODEBANC\", t = \"ABC\" Output: \"BANC\" Explain: The minimum window substring \"BANC\" includes 'A', 'B', \u0026 'C' from string t. 2๏ธโƒฃ Input: s = \"a\", t = \"a\" Output: \"a\" 3๏ธโƒฃ Input: s = \"a\", t = \"aa\" Output: \"\" Explain: Both 'a's from t must be included in the window. Since the largest window of s only has one 'a', return empty string.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n^2) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐ŸชŸ Sliding Window Pattern โฐ: O(n) ๐Ÿช: O(n)\n\n var minWindow = function (s, t) {\n let left = 0;\n let right = 0;\n let map = new Map();\n let min = Infinity;\n let result = \"\";\n for (let i = 0; i \u003c t.length; i++) {\n map.set(t[i], (map.get(t[i]) || 0) + 1);\n }\n let count = map.size;\n while (right \u003c s.length) {\n let char = s[right];\n if (map.has(char)) {\n map.set(char, map.get(char) - 1);\n if (map.get(char) === 0) {\n count--;\n }\n }\n right++;\n while (count === 0) {\n if (right - left \u003c min) {\n min = right - left;\n result = s.substring(left, right);\n }\n let char = s[left];\n if (map.has(char)) {\n map.set(char, map.get(char) + 1);\n if (map.get(char) \u003e 0) {\n count++;\n }\n }\n left++;\n }\n }\n return result;\n };\n\u003c/details\u003e\n\u003c/details\u003e\n \u003cdetails\u003e\n \u003csummary\u003e \u003ch1\u003e๐Ÿ‘ฏ Two Pointers: ๐Ÿ“Œ๐Ÿ“Œ๐ŸŽก\u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n\u003ctable\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e โ“ FIND A PAIR, TRIPLET or a SUBARRAY \u003ch3\u003e\n \u003ch3\u003eโฐ: O(n) ๐Ÿช: O(n)\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"2\"\u003e\n ๐Ÿฃ Squaring a Sorted Array, Triplets that Sum to Zero, Triplet Sum Close to Target, Triplets with Smaller Sum, Subarrays with Product Less than a Target, Comparing Strings containing Backspaces, etc.\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n\u003cpre class=\"notranslate\"\u003e\n\u003ccode\u003e\nleft = 0;\nright = 0;\nwhile (right \u003c s.length()) {\n // add s[right] to window\n right++;\n while (window needs shrink) {\n // remove s[left] from window\n left++;\n }\n}\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e\n \u003cimg\n src=\"https://cdn.emre.me/2019-10-21-two-pointers.png\"\n alt=\"Two Pointers\"\n /\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n \u003cdetails\u003e\n \u003csummary\u003e \u003ch1\u003e๐Ÿฐ๐Ÿข Fast \u0026 Slow Pointers: ๐Ÿ“Œ๐Ÿ“Œ ๐ŸŽก\u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n\u003ctable\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e โ“ FIND CYCLE IN A LINKED LIST OR FIND THE MIDDLE OF A LINKED LIST \u003c/h3\u003e\n \u003ch3\u003eโฐ:โฐ: O(n) ๐Ÿช: O(n)\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"2\"\u003e\n ๐Ÿฃ Linked List Cycle, Palindrome LinkedList, Cycle in a Circular Array, etc.\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCode\u003c/h3\u003e\n\u003cpre class=\"notranslate\"\u003e\n\u003ccode\u003e\nslow = 0;\nfast = 0;\nwhile (fast \u003c s.length()) {\n if (s[fast] is not a duplicate) {\n // move slow pointer one step\n // add s[fast] to window\n slow++;\n }\n // move fast pointer one step\n fast++;\n}\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e\n \u003cimg\n src=\"http://1algo1week.warriorkitty.com/assets/floyds-cycle-finding-algorithm/algorithm.png\"\n alt=\"Fast \u0026 Slow Pointers Pattern\"\n /\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n\u003c/details\u003e\n \u003cdetails\u003e\n \u003csummary\u003e \u003ch1\u003e๐Ÿš—๐Ÿš™ Merge Intervals : ๐Ÿ“Œ๐Ÿ“Œ๐Ÿšฅ \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n\u003ctable\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e โ“ MERGE INTERVALS OR FIND OVERLAPPING INTERVALS \u003ch3\u003e\n \u003ch3\u003e โฐ: O(nlogn) ๐Ÿช: O(n) \u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"2\"\u003e\n ๐Ÿฃ Insert Interval, Intervals Intersection, Conflicting Appointments, Minimum Meeting Rooms, etc.\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCode\u003c/h3\u003e\n\u003cpre class=\"notranslate\"\u003e\n\u003ccode\u003e\nfunction merge(intervals) {\n if (intervals.length \u003c 2) {\n return intervals;\n }\n // sort the intervals on the start time\n intervals.sort((a, b) =\u003e a[0] - b[0]);\n const mergedIntervals = [];\n let start = intervals[0][0],\n end = intervals[0][1];\n for (i = 1; i \u003c intervals.length; i++) {\n const interval = intervals[i];\n if (interval[0] \u003c= end) {\n end = Math.max(interval[1], end);\n } else {\n mergedIntervals.push([start, end]);\n start = interval[0];\n end = interval[1];\n }\n }\n // add the last interval\n mergedIntervals.push([start, end]); \n return mergedIntervals;\n}\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e\n \u003cimg\n src=\"https://cdn.emre.me/2019-10-27-merge-intervals.png\"\n alt=\"Merge Intervals Pattern\"\n /\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n\u003cdetails\u003e\n\u003csummary\u003e \u003ch1 align=\"center\"\u003e\n ๐Ÿ๐Ÿ”š Mod Binary Search: ๐Ÿ“Œ๐Ÿ“Œ๐Ÿ“Œ๐ŸŽก \n \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n\u003ctable\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003eโ“ MINIMUM DIFFERENCE OR FIND ELEMENT IN INFINITE SORTED ARRAY\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐Ÿฃ Order-agnostic Binary Search, Ceiling of a Number, Floor of a Number, Next Letter, Number Range, etc.\u003c/h3\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCode ๐Ÿ๐Ÿ”š Mod Binary Search Pattern ๐Ÿ๐Ÿ”š\u003c/h3\u003e\n\u003cpre\u003e\n\u003ccode\u003e\n start = 0, end = arr.length - 1;\n while (start \u003c= end) {\n // calculate the middle of the current range\n mid = start + (end - star\n if (key \u003c arr[mid]) {\n end = mid - 1; // the 'key' can be in the first half\n } else if (key \u003e arr[mid]) {\n start = mid + 1; // the 'key' can be in the second half\n } else { // found the key\n return mid;\n }\n }\n // element is not found\n return -1;\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e \u003cimg src=\"https://cdn.emre.me/2019-08-08-binary-search.png\"\u003e \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n \n\u003cdetails\u003e\n\u003csummary\u003e \u003ch1\u003e\u003calign=\"center\"\u003e ๐ŸŒ€ Cyclic Sort : ๐ŸŽก๐Ÿคนโ€โ™€๏ธ \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n \u003ctable\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e โ“ FIND MISSING # OR SORT #s IN PLACE \u003c/h3\u003e\n \u003ch3\u003e ๐Ÿฃ Cyclic Sort, Find the Missing Number, Find all Missing Numbers, \u003cbr/\u003e\n Find the Duplicate Number, Find all Duplicate Numbers, Find the Corrupt Pair, etc.\n \u003c/h3\u003e\n \u003ch3\u003e โฐ: O(n) ๐Ÿช: O(1) \u003c/h3\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCode\u003c/h3\u003e\n\u003cpre class=\"notranslate\"\u003e\n\u003ccode\u003e\n\ni = 0;\nwhile (i \u003c nums.length) {\n j = nums[i] - 1;\n if (nums[i] != nums[j])\n swap(nums, i, j); // put the number in its correct place\n else\n i++;\n}\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e\n \u003cimg\n src=\"https://cdn.emre.me/2019-10-28-missing-number-example.png\"\n alt=\"Cyclic Sort Pattern\"\n /\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n\u003cdetails\u003e\n\u003csummary\u003e \u003ch1\u003e๐Ÿ”€๐Ÿ”— Reverse LinkList : ๐ŸŒฑ๐Ÿคนโ€โ™€๏ธ๐ŸŽก \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n \u003ctable\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003eโ“ REVERSE A LINKEDLIST\u003c/h3\u003e\n \u003ch3\u003eโฐ: O(n) ๐Ÿช: O(1)\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐Ÿฃ Reverse a LinkedList, Reverse a Sub-list, Reverse every K-element Sub-list (medium), etc.\u003c/h3\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCode\u003c/h3\u003e\n\u003cpre\u003e\n\u003ccode\u003e\n function reverse(head) {\n let prev = null;\n while (head !== null) {\n next = head.next;\n head.next = prev;\n prev = head;\n head = next;\n }\n return prev;\n }\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e \u003cimg src=\"https://cdn.emre.me/2019-11-04-in-place-reversal.gif\"\u003e \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n\u003cdetails\u003e\n\u003csummary\u003e \u003ch1 align=\"center\"\u003e\n ๐ŸŒณ BFS : ๐Ÿฆœ ๐Ÿชž\n \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n\u003ctable style=\"width:100%\"\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003eโ“ FIND MIN DEPTH, MAX DEPTH, LEVEL AVERAGE OF BINARY TREE\u003c/h3\u003e\n \u003ch3\u003eโฐ: O(n) ๐Ÿช: O(n)\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐Ÿฃ Level Order Traversal, Zigzag Traversal, Level Averages in a Binary Tree, \u003cbr/\u003e \n Minimum Depth of a Binary Tree, Level Order Successor, \u003cbr/\u003e\n Connect Level Order Siblings, etc. Tree Breadth First Search?\u003c/h3\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCode ๐ŸŒณ\u003c/h3\u003e\n\u003cpre\u003e\n\u003ccode\u003e\n const queue = [root];\n while (queue.length) {\n const currentNode = queue.shift();\n if (currentNode.left) queue.push(currentNode.left);\n if (currentNode.right) queue.push(currentNode.right);\n }\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e \u003cimg src=\"https://www.guru99.com/images/1/020820_0543_BreadthFirs1.png\"\u003e \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n\u003cdetails\u003e\n\u003csummary\u003e \u003ch1 align=\"center\"\u003e\n ๐ŸŒฒ DFS : ๐Ÿฝ๐Ÿชž\n \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n\u003ctable style=\"width:100%\"\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003eโ“ FIND PATH WITH MAX SUM OR PATH WITH GIVEN SEQUENCE\u003c/h3\u003e\n \u003ch3\u003eโฐ: O(n) ๐Ÿช: O(n)\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐Ÿฃ Reverse Level Order Traversal, Zigzag Traversal, \u003cbr/\u003e\n Level Averages in a Binary Tree, Minimum Depth of a Binary Tree, \u003cbr/\u003e\n Level Order Successor, Connect Level Order Siblings, etc.\u003c/h3\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCode Tree Depth First Search Pattern ๐ŸŒฒ\u003c/h3\u003e\n\u003cpre\u003e\n\u003ccode\u003e\n Stack\u003c Tree Node stack = new Stack\u003c\u003e();\n stack.push(root);\n while (!stack.isEmpty()) {\n TreeNode currentNode = stack.pop(); \n if (currentNode.left != null) stack.push(currentNode.left);\n if (currentNode.right != null) stack.push(currentNode.right);\n }\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e \u003cimg src=\"https://assets.leetcode.com/users/andvary/image_1556551007.png\"\u003e \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n\u003cdetails\u003e\n\u003csummary\u003e \u003ch1 align=\"center\"\u003e\n ๐Ÿ“ฆ๐Ÿ“ฆ Two Heaps : ๐Ÿ“ฆ๐Ÿ“ฆ ๐ŸŽก\n \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n\u003ctable\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003eโ“ MEDIAN OF # STREAM FIND K SMALLEST #\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐Ÿฃ Sliding Window Median, Find the Median of a Number Stream, etc.\u003c/h3\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCode Two Heaps Pattern ๐Ÿ“ฆ๐Ÿ“ฆ\u003c/h3\u003e\n\u003cpre\u003e\n\u003ccode\u003e\n let maxHeap = new MaxHeap();\n let minHeap = new MinHeap();\n for (num of nums) {\n if (maxHeap.isEmpty() || num \u003c= maxHeap.peek()) {\n maxHeap.push(num);\n } else {\n minHeap.push(num);\n }\n if (maxHeap.size() \u003e minHeap.size() + 1) {\n minHeap.push(maxHeap.pop());\n } else if (maxHeap.size() \u003c minHeap.size()) {\n maxHeap.push(minHeap.pop());\n }\n }\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e \u003cimg src=\"https://media.geeksforgeeks.org/wp-content/cdn-uploads/MinHeapAndMaxHeap.png\"\u003e \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n\u003cdetails\u003e\n\u003csummary\u003e \u003ch1 align=\"center\"\u003e\n ๐Ÿ› Subsets : ๐Ÿšฅ๐ŸŽก\n \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n\u003ctable\u003e \n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003eโ“ FIND ALL SUBSETS OF A SET OR FIND ALL SUBSETS ADD UP TO GIVEN #PERMUTATIONS AND COMBINATIONS OF SUBSETS\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐Ÿฃ Find all subsets of a set, Find all subsets of a set with duplicates, Find all subsets with a given sum, etc.\u003c/h3\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCode \u003c/h3\u003e\n\u003cpre\u003e\n\u003ccode\u003e\n let subsets = [[]];\n for (let i = 0; i \u003c nums.length; i++) {\n let n = subsets.length;\n for (let j = 0; j \u003c n; j++) {\n let set = subsets[j].slice(0);\n set.push(nums[i]);\n subsets.push(set);\n }\n }\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e \u003cimg src=\"https://hackernoon.com/images/G9YRlqC9joZNTWsi1ul7tRkO6tv1-hemg3w8d.jpg\"\u003e \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n\n\u003cdetails\u003e\n\u003csummary\u003e \u003ch1 align=\"center\"\u003e\n #๏ธโƒฃ ๐Ÿ‘‘ K TOP : ๐Ÿ“ฆ ๐ŸŽก๐ŸŽก\n \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n\u003ctable\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003eโ“ TOP K #s OR FREQUENCY OF TOP K #s\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐Ÿฃ Top 'K' Numbers, Kth Largest Number in a Stream, K Closest Points to the Origin, etc.\u003c/h3\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCode K TOP in Javascript:\u003c/h3\u003e\n\u003cpre\u003e\n\u003ccode\u003e\n import MinHeap from './DataStructures/Heap/MinHeap.js';\n\n function findLargestKNum(nums, k) {\n const minHeap = new MinHeap();\n for (i = 0; i \u003c k; i++) {\n minHeap.push(nums[i]);\n }\n for (i = k; i \u003c nums.length; i++) {\n if (nums[i] \u003e minHeap.peek()) {\n minHeap.pop();\n minHeap.push(nums[i]);\n }\n }\n\n return minHeap.toArray();\n }\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e \u003cimg src=\"https://i.ytimg.com/vi/Wh3A29psE_Y/maxresdefault.jpg\"\u003e \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n\u003cdetails\u003e\n\u003csummary\u003e \u003ch1 align=\"center\"\u003e\n #๏ธโƒฃ๐Ÿš•๐Ÿš“ K MERGE : ๐ŸŒฑ๐Ÿ“ฆ๐ŸŽก๐ŸŽก\n \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n\u003ctable\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003eโ“ MERGE K SORTED ARRAYS OR MERGE K SORTED LISTS\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐Ÿฃ Merge K Sorted Lists, Kth Smallest Number in M Sorted Lists, Kth Smallest Number in a Sorted Matrix, etc.\u003c/h3\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCodeK MERGE Pattern in Javascript:\u003c/h3\u003e\n\u003cpre\u003e\n\u003ccode\u003e\n import ListNode from \"DataStructures/LinkedList/ListNode.js\";\n import MinHeap from \"DataStructures/Heaps/MinHeap.js\";\n\n const mergeKLists = function (lists) {\n if (lists.length === 0) return null;\n let dummy = new ListNode();\n let curr = dummy;\n let minHeap = new MinHeap();\n for (let i = 0; i \u003c lists.length; i++) {\n if (lists[i]) minHeap.insert(lists[i]);\n }\n while (minHeap.size() \u003e 0) {\n let node = minHeap.remove();\n curr.next = node;\n curr = curr.next;\n if (node.next) minHeap.insert(node.next);\n }\n return dummy.next;\n };\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003ctd colspan=\"3\"\u003e \u003cimg src=\"https://i.ytimg.com/vi/Xo54nlPHSpg/maxresdefault.jpg\"\u003e \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n\u003cdetails\u003e\n\u003csummary\u003e \u003ch1 align=\"center\"\u003e\n ๐Ÿ“… Topological Sort : ๐Ÿ“ฆ๐Ÿ“ฑ๐ŸŽก\n \u003c/h1\u003e \u003c/summary\u003e\n \u003cbr\u003e\n\u003ctable\u003e \n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003eโ“ FIND ORDER OF TASKS OR IF GIVEN SEQUENCE IS VALID\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd\u003e\n \u003cimg src=\"https://slideplayer.com/slide/12886082/78/images/5/Topological+Sort%3A+Definition.jpg\" alt=\"Topological Sort: Definition\" width=\"500\" height=\"500\"\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n \u003ctr\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐Ÿฃ Tasks Scheduling, Tasks Scheduling Order, All Tasks Scheduling Orders, etc.\u003c/h3\u003e\n \u003c/td\u003e\n \u003ctd\u003e\n \u003ch3\u003e๐ŸŽญ PsuendoCode ๐Ÿ“… Topological Sort Pattern ๐Ÿ“… in Javascript:\u003c/h3\u003e\n\u003cpre\u003e\n\u003ccode\u003e\nconst Deque = require('collections/deque'); //http://www.collectionsjs\nfunction print_orders(tasks, prerequisites) {\n sortedOrder = [];\n if (tasks \u003c= 0) {\n return false;\n\n // a. Initialize the graph\n inDegree = Array(tasks).fill(0); // count of incoming edges\n graph = Array(tasks).fill(0).map(() =\u003e Array()); // adjacency list g\n // b. Build the graph\n prerequisites.forEach((prerequisite) =\u003e {\n let parent = prerequisite[0],\n child = prerequisite[1];\n graph[parent].push(child); // put the child into it's parent's list\n inDegree[child]++; // increment child's inDegree\n\n // c. Find all sources i.e., all vertices with 0 in-degrees\n sources = new Deque();\n for (i = 0; i \u003c inDegree.length; i++) {\n if (inDegree[i] === 0) {\n sources.push(i);\n }\n\n print_all_topological_sorts(graph, inDegree, sources, sortedOrder);\n return sortedOrder;\n\nfunction print_all_topological_sorts(graph, inDegree, sources, sortedOrder) {\n if (sources.length \u003e 0) {\n for (i = 0; i \u003c sources.length; i++) {\n vertex = sources.shift();\n sortedOrder.push(vertex);\n sourcesForNextCall = sources.slice(0); // make a copy of sources\n // only remove the edges, if all of its children are not sources\n graph[vertex].forEach((child) =\u003e { // get the node's children to decrement their in-degrees\n inDegree[child]--; // decrement inDegree of child\n if (inDegree[child] === 0) {\n sourcesForNextCall.push(child); // save the new source for the next call\n }\n });\n // recursive call to print other orderings from the remaining (and new) sources\n print_all_topological_sorts(graph, inDegree, sourcesForNextCall, sortedOrder);\n // backtrack, remove the vertex from the sorted order and put all of its children back to consider\n // the next source instead of the current vertex\n sortedOrder\n\u003c/code\u003e\n\u003c/pre\u003e\n \u003c/td\u003e\n \u003c/tr\u003e\n\u003c/table\u003e\n\u003c/details\u003e\n\n\n\n \n\n\n \n\n\n\n\n\n\n## Other Patterns\n\n## Union Find Algorithm Pattern โ™พ -\u003e\n โ“ # OF CONNECTED COMPONENETS IN UNDIRECTED GRAPH, FIND IF 2 NODES CONNECTED\n ๐Ÿฃ Number of Connected Components in an Undirected Graph, Find whether two nodes are connected in an undirected graph, etc.\n\n ๐ŸŽญ PsuendoCode Union Find Algorithm Pattern โ™พ\n โฐ: O(V * logV) ๐Ÿช: O\n function find(int[] parent, i) {\n if (parent[i] == -1) return i;\n return find(parent, parent[i]);\n \n function union(int[] parent, x, y) {\n xset = find(parent, x);\n yset = find(parent, y);\n parent[xset] = yset;\n }\n## Greedy Pattern ๐Ÿ’ฐ -\u003e\n โ“ MINIMUM # OF MEETINGS MAX OF INTERVALS NOT OVERLAPPING\n ๐Ÿฃ Activity Selection Problem, Coin Change, Fractional Knapsack Problem, Job Sequencing Problem, Huffman Coding, Prim's Minimum Spanning Tree, Kruskal's Minimum Spanning Tree, Dijkstra's Shortest Path Algorithm, Bellman-Ford Algorithm, Floyd-Warshall Algorithm, Travelling Salesman Problem\n\n ๐ŸŽญ PsuendoCode Greedy Pattern ๐Ÿ’ฐ\n โฐ: O(nlogn) ๐Ÿช: O(1)\n \n Arrays.sort(intervals, (a, b) -\u003e Integer.compare(a[0], b[0]));\n count = 0, end = Integer.MIN_VALUE;\n for (int[] interval : intervals) {\n if (interval[0] \u003e= end) {\n count++;\n end = interval[1];\n }\n }\n return count;\n![alt text](https://www.quizzcreator.com/img/QUIZ/quiz/60b2236b8c9666.00493324.173.2077.jpg);\n## Backtracking Pattern ๐ŸŽฒ -\u003e\n โ“ FIND ALL PERMUTATIONS COMBINATIONS SUBSETS PARTIONING \n ๐Ÿฃ N-Queens Problem, Sudoku Solver, Rat in a Maze, Knight's Tour Problem, Hamiltonian Cycle, Subset Sum Problem, Permutations, Combination Sum, Palindrome Partitioning, Word Break Problem, etc.\n\n ๐ŸŽญ PsuendoCode Backtracking Pattern ๐ŸŽฒ\n โฐ: O(n!) ๐Ÿช: O(n)\n \n function backtrack(n, ArrayList\u003cInteger\u003e nums, List\u003cList\u003cInteger\u003e\u003e output, first) {\n // if all integers are used up\n if (first == n)\n output.add(new ArrayList\u003cInteger\u003e(nums));\n for (i = first; i \u003c n; i++) {\n // place i-th integer first \n // in the current permutation\n Collections.swap(nums, first, i);\n // use next integers to complete the permutations\n backtrack(n, nums, output, first + 1);\n // backtrack\n Collections.swap(nums, first, i);\n }\n![alt text](https://ibpublicimages.s3-us-west-2.amazonaws.com/tutorial/backtracking1.png);\n## Dynamic Programming Pattern ๐Ÿ“ˆ -\u003e\n โ“ FIND OPTIMAL SOLUTION TO COMPLEX PROBLEMS MIN MAX OR COUNT OF GIVEN CONSTRAINTS\n ๐Ÿฃ Fibonacci Numbers, House Thief, Minimum Coin Change, House Painters, Palindromic Subsequence, Longest Common Subsequence, Longest Increasing Subsequence, Longest Common Substring, Edit Distance, 0/1 Knapsack Problem, Subset Sum Problem, Unbounded Knapsack Problem, Rod Cutting, Word Break Problem, etc.\n\n ๐ŸŽญ PsuendoCode Dynamic Programming Pattern ๐Ÿ“ˆ\n โฐ: O(n) ๐Ÿช: O(n)\n \n int[] dp = new int[n + 1];\n dp[0] = 1;\n dp[1] = 1;\n for (i = 2; i \u003c= n; i++)\n dp[i] = dp[i - 1] + dp[i - 2];\n return dp[n];\n![alt text](https://imgs.developpaper.com/imgs/51946-20190827144809463-5507841.png);\n## ๐Ÿงฉ Bit Manipulation Pattern ๐Ÿงฉ -\u003e\n 1๏ธโƒฃ Bitwise XOR\n โ“ FIND IF NUMBER IS POWER OF 2\n ๐Ÿฃ Power of 2, Bitwise AND of Numbers Range, etc.\n\n ๐ŸŽญ PsuendoCode ๐Ÿงฉ Bit Manipulation Pattern ๐Ÿงฉ\n โฐ: O(1) ๐Ÿช: O(1)\n return (n \u0026 (n - 1)) == 0;\n\n โ“ FIND UNIQUE NUMBER IN ARRAY OF PAIRS\n ๐Ÿฃ Single Number, Find the Missing Number, Find the Duplicate Number, Find the Corrupt Pair, etc.\n 2๏ธโƒฃ Bitwise AND\n ๐ŸŽญ PsuendoCode Bitwise XOR Pattern ๐Ÿงฉ\n โฐ: O(n) ๐Ÿช: O(1)\n int n = nums.length;\n for (i = 0; i \u003c n; i++)\n x1 = x1 ^ nums[i];\n for (i = 1; i \u003c= n + 1; i++)\n x2 = x2 ^ i;\n return x1 ^ x2;\n 3๏ธโƒฃ Bitwise OR\n โ“ FIND IF NUMBER IS POWER OF 4\n ๐Ÿฃ Power of 4, etc.\n ๐ŸŽญ PsuendoCode Bitwise OR Pattern ๐Ÿงฉ\n โฐ: O(1) ๐Ÿช: O(1)\n return (n \u003e 0) \u0026\u0026 ((n \u0026 (n - 1)) == 0) \u0026\u0026 ((n \u0026 0xAAAAAAAA) == 0);\n\n![alt text](https://miro.medium.com/max/620/1*CxPeaIX4QxILRzucksVCsw.png);\n\n 4๏ธโƒฃ Bitwise Left Shift\n โ“ FIND IF NUMBER IS POWER OF 2\n ๐Ÿฃ Power of 2, etc.\n ๐ŸŽญ PsuendoCode Bitwise Left Shift Pattern ๐Ÿงฉ\n โฐ: O(1) ๐Ÿช: O(1)\n return (n \u003e 0) \u0026\u0026 ((n \u0026 (n - 1)) == 0);\n\n![alt text](https://cdn.educba.com/academy/wp-content/uploads/2020/05/Left-Shift-Operator-in-C-01.png);\n \n 5๏ธโƒฃ Bitwise Right Shift\n โ“ FIND IF NUMBER IS POWER OF 2\n ๐Ÿฃ Power of 2, etc.\n ๐ŸŽญ PsuendoCode Bitwise Right Shift Pattern ๐Ÿงฉ\n โฐ: O(1) ๐Ÿช: O(1)\n return (n \u003e 0) \u0026\u0026 ((n \u0026 (n - 1)) == 0);\n![alt text](https://he-s3.s3.amazonaws.com/media/uploads/5f05f51.jpg);\n## ๐Ÿ“ Matrix Pattern ๐Ÿ“ -\u003e\n โ“ FIND IF MATRIX IS SINGULAR\n ๐Ÿฃ Matrix Chain Multiplication, etc.\n\n ๐ŸŽญ PsuendoCode ๐Ÿ“ Matrix Pattern ๐Ÿ“\n โฐ: O(n^3) ๐Ÿช: O(n^2)\n int n = matrix.length;\n for (i = 0; i \u003c n; i++) {\n if (matrix[i][i] == 0) {\n boolean zeroRow = true;\n for (j = 0; j \u003c n; j++) {\n if (matrix[i][j] != 0) {\n zeroRow = false;\n break;\n }\n }\n if (zeroRow) return true;\n boolean zeroColumn = true;\n for (j = 0; j \u003c n; j++) {\n if (matrix[j][i] != 0) {\n zeroColumn = false;\n break;\n }\n }\n if (zeroColumn) return true;\n }\n }\n return false;\n![alt text](https://www.geeksforgeeks.org/wp-content/uploads/strassen_new.png);\n## ๐Ÿ“˜ Trie Pattern ๐Ÿ“˜ -\u003e\n โ“ FIND ALL WORDS WITH GIVEN PREFIX\n ๐Ÿฃ Word Search, Word Break Problem, etc.\n\n ๐ŸŽญ PsuendoCode ๐Ÿ“˜ Trie Pattern ๐Ÿ“˜\n โฐ: O(n) ๐Ÿช: O(n)\n TrieNode root = new TrieNode();\n for (String word : words) {\n TrieNode node = root;\n for (char letter : word.toCharArray()) {\n if (node.children[letter - 'a'] == null)\n node.children[letter - 'a'] = new TrieNode();\n node = node.children[letter - 'a'];\n }\n node.word = word;\n }\n return root;\n![alt text](https://miro.medium.com/max/630/1*f89l46VKjKPgnecqOw4V_w.png);\n\n\n# 75 BLIND CURATED LEETCODE QUESTIONS: \n## Array\n## [Two Sum](https://leetcode.com/problems/two-sum/) #1 ๐Ÿ‘ฏ \n โ“: Given an array of integers nums \u0026 an integer target, return indices of the two numbers such that they add up to target.\n ๐Ÿฃ: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N^2) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ‘ฏ Two Pointers โฐ: O(NlogN) ๐Ÿช: O(1)\n\n var twoSum = function(nums, target) {\n var twoSum = function(nums, target) {\n let checkSum = new Map();\n for(let i=0; i\u003cnums.length; i++){\n var curr = nums[i];\n let diff = target-nums[i];\n if(checkSum.get(diff)){\n if(checkSum.get(diff)[0] === curr) {\n return [checkSum.get(diff)[1], i ];\n } \n } else {\n checkSum.set(nums[i], [diff, i]);\n }\n }\n return checkSum\n };\n\n## [Best Time to Buy \u0026 Sell Stock](https://leetcode.com/problems/best-time-to-buy-\u0026-sell-stock/) #121 ๐Ÿ“ˆ\n โ“: You are given an array prices where prices[i] is the price of a given stock on the ith day.\n You want to maximize your profit by choosing a single day to buy one stock \u0026 choosing a different day in the future to sell that stock.\n Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.\n ๐Ÿฃ: Input: prices = [7,1,5,3,6,4], Output: 5, Explain: Buy on day 2 (price = 1) \u0026 sell on day 5 (price = 6), profit = 6-1 = 5. Not 7-1 = 6, as selling price needs to be larger than buying price.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N^2) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Pattern Kadane's Algorithm โฐ: O(N) ๐Ÿช: O(1)\n\n var maxProfit = function(prices) {\n let min = prices[0];\n let max = 0;\n for(let i=1; i\u003cprices.length; i++){\n if(prices[i] \u003c min){\n min = prices[i];\n } else {\n max = Math.max(max, prices[i]-min);\n }\n }\n return max;\n };\n\n## [Contains Duplicate](https://leetcode.com/problems/contains-duplicate/) ๐Ÿ‘ฏ \n Problem : Given Arr [] of int. Return true if has duplicates : \n Parameters: Ex 1: Input: nums = [1,2,3,1] Output: true. Ex 2: Input: nums = [1,2,3,4] Output: false. \n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N^2) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ‘ฏ Two Pointers โฐ: O(N) ๐Ÿช: O(1)\n\n var containsDuplicate = function(nums) {\n let checkSum = new Map();\n for(let i=0; i\u003cnums.length; i++){\n if(checkSum.get(nums[i])){\n return true;\n } else {\n checkSum.set(nums[i], 1);\n }\n }\n return false;\n };\n\n var containsDuplicate = function(nums) {\n return [...new Set(nums)].length !== nums.length;\n }; \n\n## [Product of Array Except Self](https://leetcode.com/problems/product-of-array-except-self/) #238 ๐Ÿ‘ฏ \n โ“: Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].\n The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.\n You must write an algorithm that runs in O(n) time \u0026 without using the division operation.\n ๐Ÿฃ: Input: nums = [1,2,3,4], Output: [24,12,8,6]\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N^2) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ‘ฏ Two Pointers โฐ: O(N) ๐Ÿช: O(1)\n \n var productExceptSelf = function(nums) {\n let result = [];\n let left = 1;\n let right = 1;\n for(let i=0; i\u003cnums.length; i++){\n result[i] = left;\n left *= nums[i];\n }\n for(let i=nums.length-1; i\u003e=0; i--){\n result[i] *= right;\n right *= nums[i];\n }\n return result;\n };\n\n## [Maximum Subarray](https://leetcode.com/problems/maximum-subarray/) #53 ๐Ÿ“ˆ\n โ“: Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum \u0026 return its sum.\n ๐Ÿฃ: Input: nums = [-2,1,-3,4,-1,2,1,-5,4], Output: 6, Explain: [4,-1,2,1] has the largest sum = 6.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N^2) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Pattern Kadane's Algorithm โฐ: O(N) ๐Ÿช: O(1)\n\n var maxSubArray = function(nums) {\n let max = nums[0];\n let curr = nums[0];\n for(let i=1; i\u003cnums.length; i++){\n curr = Math.max(nums[i], curr+nums[i]);\n max = Math.max(max, curr);\n }\n return max;\n };\n\n## [Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/) #152 ๐Ÿ“ˆ\n โ“: Given an integer array nums, find a contiguous non-empty subarray within the array that has the largest product, \u0026 return the product.\n It is guaranteed that the answer will fit in a 32-bit integer.\n A subarray is a contiguous subsequence of the array.\n ๐Ÿฃ: Input: nums = [2,3,-2,4], Output: 6, Explain: [2,3] has the largest product 6.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N^2) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Pattern Kadane's Algorithm โฐ: O(N) ๐Ÿช: O(1)\n\n var maxProduct = function(nums) {\n let max = nums[0];\n let min = nums[0];\n let result = nums[0];\n for(let i=1; i\u003cnums.length; i++){\n let temp = max;\n max = Math.max(nums[i], Math.max(max*nums[i], min*nums[i]));\n min = Math.min(nums[i], Math.min(temp*nums[i], min*nums[i]));\n result = Math.max(result, max);\n }\n return result;\n };\n## [Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) #153 ๐Ÿ๐Ÿ”š\n โ“: Suppose an array of length n sorted in ascending order is rotated between 1 \u0026 n times. For example, the array nums = [0,1,2,4,5,6,7] might become:\n [4,5,6,7,0,1,2] if it was rotated 4 times.\n [0,1,2,4,5,6,7] if it was rotated 7 times.\n Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].\n Given the sorted rotated array nums of unique elements, return the minimum element of this array.\n ๐Ÿฃ: Input: nums = [3,4,5,1,2], Output: 1\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ๐Ÿ”š Mod Binary Search Pattern โฐ: O(logN) ๐Ÿช: O(1)\n\n Pattern: ๐Ÿ๐Ÿ”š Mod Binary Search\n var findMin = function(nums) {\n let left = 0;\n let right = nums.length-1;\n while(left \u003c right){\n let mid = Math.floor((left+right)/2);\n if(nums[mid] \u003e nums[right]){\n left = mid+1;\n } else {\n right = mid;\n }\n }\n return nums[left];\n };\n## [Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) #153 ๐Ÿ๐Ÿ”š \n โ“: Suppose an array of length n sorted in ascending order is rotated between 1 \u0026 n times. For example, the array nums = [0,1,2,4,5,6,7] might become:\n [4,5,6,7,0,1,2] if it was rotated 4 times.\n [0,1,2,4,5,6,7] if it was rotated 7 times.\n Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].\n Given the sorted rotated array nums of unique elements, return the minimum element of this array.\n ๐Ÿฃ: Input: nums = [3,4,5,1,2], Output: 1\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ๐Ÿ”š Mod Binary Search โฐ: O(logN) ๐Ÿช: O(1)\n\n var findMin = function(nums) {\n let left = 0;\n let right = nums.length-1;\n while(left \u003c right){\n let mid = Math.floor((left+right)/2);\n if(nums[mid] \u003e nums[right]){\n left = mid+1;\n } else {\n right = mid;\n }\n }\n return nums[left];\n };\n## [Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/) #33 ๐Ÿ๐Ÿ”š\n โ“: You are given an integer array nums sorted in ascending order, \u0026 an integer target.\n Suppose that nums is rotated at some pivot unknown to you beforeh\u0026 (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).\n If target is found in the array return its index, otherwise, return -1.\n ๐Ÿฃ: Input: nums = [4,5,6,7,0,1,2], target = 0, Output: 4\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ๐Ÿ”š Mod Binary Search โฐ: O(logn) ๐Ÿช: O(1) \n \n var search = function(nums, target) {\n let left = 0;\n let right = nums.length-1;\n while(left \u003c= right){\n let mid = Math.floor((left+right)/2);\n if(nums[mid] === target){\n return mid;\n } else if(nums[mid] \u003e= nums[left]){\n if(target \u003e= nums[left] \u0026\u0026 target \u003c nums[mid]){\n right = mid-1;\n } else {\n left = mid+1;\n }\n } else {\n if(target \u003e nums[mid] \u0026\u0026 target \u003c= nums[right]){\n left = mid+1;\n } else {\n right = mid-1;\n }\n }\n }\n return -1;\n };\n## [3Sum](https://leetcode.com/problems/3sum/) #15 ๐Ÿ‘ฏ\n โ“: Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, \u0026 j != k, \u0026 nums[i] + nums[j] + nums[k] == 0.\n Notice that the solution set must not contain duplicate triplets.\n ๐Ÿฃ: Input: nums = [-1,0,1,2,-1,-4], Output: [[-1,-1,2],[-1,0,1]]\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N^3) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ‘ฏ Two Pointers โฐ: O(N^2) ๐Ÿช: O(N)\n\n var threeSum = function(nums) {\n let result = [];\n nums.sort((a,b) =\u003e a-b);\n for(let i=0; i\u003cnums.length-2; i++){\n if(i \u003e 0 \u0026\u0026 nums[i] === nums[i-1]) continue;\n let left = i+1;\n let right = nums.length-1;\n while(left \u003c right){\n let sum = nums[i] + nums[left] + nums[right];\n if(sum === 0){\n result.push([nums[i], nums[left], nums[right]]);\n while(left \u003c right \u0026\u0026 nums[left] === nums[left+1]) left++;\n while(left \u003c right \u0026\u0026 nums[right] === nums[right-1]) right--;\n left++;\n right--;\n } else if(sum \u003c 0){\n left++;\n } else {\n right--;\n }\n }\n }\n return result;\n };\n## [Container With Most Water](https://leetcode.com/problems/container-with-most-water/) #11 ๐Ÿ‘ฏ \n โ“: Given n non-negative integers a1, a2, ..., an , where each represents a poat coordinate (i, ai) n vertical lines are drawn such that the two endpoints of the line i is at (i, ai) \u0026 (i, 0) Find two lines, which, together with the x-axis forms a container, such that the container contains the most water\n Notice that you may not slant the container.\n ๐Ÿฃ: Input: height = [1,8,6,2,5,4,8,3,7], Output: 49\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n^2) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ‘ฏ Two Pointers โฐ: O(n) ๐Ÿช: O(1)\n\n var maxArea = function(height) {\n let left = 0;\n let right = height.length-1;\n let max = 0;\n while(left \u003c right){\n let area = Math.min(height[left], height[right]) * (right-left);\n max = Math.max(max, area);\n if(height[left] \u003c height[right]){\n left++;\n } else {\n right--;\n }\n }\n return max;\n };\n\n---\n\n## Binary \n\n## [Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/) ๐Ÿงฉ \n โ“: Given two integers a \u0026 b, return sum of the two integers without using the operators + \u0026 -.\n ๐Ÿฃ: 1๏ธโƒฃ Input: a = 1, b = 2. Output: 3 2๏ธโƒฃ Input: a = 2, b = 3 Output: 5.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿงฉ Bit Manipulation โฐ: O(1) ๐Ÿช: O(1)\n\n var getSum = function(a, b) {\n let carry; // store carry \n while(a!==0){\n carry = a \u0026 b;\n b = a ^ b;\n a = carry \u003c\u003c 1;\n }\n return b \n };\n## [Divide Two Integers](https://leetcode.com/problems/divide-two-integers/) #29 ๐Ÿงฉ \n โ“: Given two integers dividend \u0026 divisor, divide two integers without using multiplication, division, \u0026 mod operator.\n Return the quotient after dividing dividend by divisor.\n The integer division should truncate toward zero, which means losing its fractional part. For example, truncate(8.345) = 8 \u0026 truncate(-2.7335) = -2.\n Note: Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [โˆ’231, 231 โˆ’ 1]. For this problem, assume that your function returns 231 โˆ’ 1 when the division result overflows.\n ๐Ÿฃ: Input: dividend = 10, divisor = 3, Output: 3\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿงฉ Bit Manipulation โฐ: O(logn) ๐Ÿช: O(logn)\n\n var divide = function(dividend, divisor) {\n if (dividend === 0 || divisor === 0) return 0\n let isNegative = false;\n let quotient = 0;\n if ((divisor \u003c 0 || dividend \u003c 0) \u0026\u0026 !(divisor \u003c 0 \u0026\u0026 dividend \u003c 0)) isNegative = true;\n dividend = Math.abs(dividend)\n divisor = Math.abs(divisor)\n while (dividend \u003e= divisor) {\n let carry = 1, tempDividend= dividend, tempDivisor = divisor\n while (tempDivisor \u003c= (tempDividend\u003e\u003e 1)){\n carry \u003c\u003c= 1\n tempDividend\u003e\u003e= 1\n tempDivisor \u003c\u003c= 1\n }\n quotient += carry\n dividend -= tempDivisor\n }\n if (isNegative) return -quotient\n if (quotient \u003e= 2**31) {\n return 2**31 - 1\n }\n return quotient\n };\n## [Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/) #191 ๐Ÿงฉ \n โ“: Write a function that takes an unsigned integer \u0026 returns the number of '1' bits it has (aka Hamming weight).\n ๐Ÿฃ: 1๏ธโƒฃ Input: n = 00000000000000000000000000001011 Output: 3 . N as a total of three '1' bits.2๏ธโƒฃ Input: n = 00000000000000000000000010000000 Output: 1 3๏ธโƒฃ Input: n = 11111111111111111111111111111101 Output: 31\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿงฉ Bit Manipulation โฐ: O(1) ๐Ÿช: O(1)\n\n var hammingWeight = function(n) {\n let count = 0;\n while(n!=0){\n n = n\u0026(n-1); //mask\u003c\u003c1\n count++;\n }\n return count\n };\n\n## [Counting Bits](https://leetcode.com/problems/counting-bits/) #338 ๐Ÿงฉ \n โ“: Given an integer n, return an array ans of length n + 1 such that for each i (0 \u003c= i \u003c= n), ans[i] is the number of 1's in the binary representation of i.\n ๐Ÿฃ: 1๏ธโƒฃ Input: n = 2 Output: [0,1,1] Explain: 0 --\u003e 0 1 --\u003e 1 2 --\u003e 10 2๏ธโƒฃ Input: n = 5 Output: [0,1,1,2,1,2]\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿงฉ Bit Manipulation โฐ: O(N) ๐Ÿช: O(1)\n\n var countBits = function(n) {\n let result = [0];\n for(let i = 1; i \u003c= n; i++){\n result.push(result[i\u003e\u003e1] + (i\u00261));\n }\n return result;\n };\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N) ๐Ÿช: O(N)\n ๐Ÿ‡ Solution: ๐Ÿงฉ Bit Manipulation โฐ: O(n) ๐Ÿช: O(n)\n\n const countBits = (n) =\u003e {\n const res = new Array(n);\n res[0] = 0;\n\n for (let i = 1; i \u003c= n; i++) {\n if (i % 2 === 0) res[i] = res[i / 2];\n else res[i] = res[i - 1] + 1;\n }\n\n return res;\n };\n## [Missing Number](https://leetcode.com/problems/missing-number/) #268 ๐Ÿงฉ\n โ“: Given an [] containing n distinct numbers in the range [0, n], return the only num in the range tht missing from the []\n ๐Ÿฃ: 1๏ธโƒฃ Input: nums = [3,0,1] Output: 2 Explain: n = 3 since there are 3 num, so all n are in range [0,3]. 2 is the missing num in the range it does not appear in nums. 2๏ธโƒฃ Input: nums = [0,1] Output: 2 3๏ธโƒฃ Input: nums = [9,6,4,2,3,5,7,0,1] Output: 8\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N) ๐Ÿช: O(N)\n ๐Ÿ‡ Solution: ๐Ÿงฉ Bit Manipulation โฐ: O(n) ๐Ÿช: O(n)\n\n var missingNumber = function (nums) {\n let i = 0;\n let temp = 0;\n while (i \u003c nums.length) {\n if (nums[i] != i \u0026\u0026 nums.length \u003e nums[i]) {\n temp = nums[nums[i]];\n nums[nums[i]] = nums[i];\n nums[i] = temp;\n } else {\n i++;\n }\n }\n\n for (let i = 0; i \u003c nums.length; i++) {\n if (i != nums[i]) return i;\n }\n\n return nums.length;\n };\n\n## [Reverse Bits](https://leetcode.com/problems/reverse-bits/) #190 ๐Ÿงฉ\n โ“: Reverse bits of a given 32 bits unsigned integer.\n ๐Ÿฃ: 1๏ธโƒฃ Input: n = 00000010100101000001111010011100 Output: 964176192 (00111001011110000010100101000000)\n Explain: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000. 2๏ธโƒฃ Input: n = 11111111111111111111111111111101\n Output: 3221225471 (10111111111111111111111111111111)\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(N) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿงฉ Bit Manipulation โฐ: O(1) ๐Ÿช: O(1)\n\n var reverseBits = function (n) {\n var len = 32;\n var arr = new Array();\n while (len \u003e 0) {\n var t = n \u0026 1;\n n = n \u003e\u003e 1;\n arr.push(t);\n len--;\n }\n var res = arr.join(\"\");\n return parseInt(res, 2);\n };\n\n## ๐Ÿ“ˆ Dynamic Programming\n\n## [Climbing Stairs](https://leetcode.com/problems/climbing-stairs/) #70 ๐Ÿ“ˆ\n โ“: You are climbing a staircase. It takes n steps to reach the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?\n ๐Ÿฃ: 1๏ธโƒฃ Input: n = 2 Output: 2. Explain: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps\n 2๏ธโƒฃ Input: n = 3 Output: 3. Explain: There are three ways to climb to the top. 1. 1 step + 1 step + 1 step 2. 1 step + 2 steps 3. 2 steps + 1 step\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(2^n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Memoization โฐ: O(n) ๐Ÿช: O(n)\n\n var climbStairs = function (n) {\n if (n == 1) return 1;\n if (n == 2) return 2;\n let dp = new Array(n + 1); // ways to climb array\n dp[1] = 1;\n dp[2] = 2;\n for (let i = 3; i \u003c= n; i++) { //Either climb 1 stair \u0026 then climb the rest i-1 stairs X ways.\n dp[i] = dp[i - 1] + dp[i - 2]; //- Or climb 2 stairs at once \u0026 then climb the rest i-2 stairs Y ways.\n } //Total: X + Y ways\n return dp[n]; \n };\n## [Coin Change](https://leetcode.com/problems/coin-change/) #322 ๐Ÿ“ˆ\n โ“: You are given an integer array coins representing coins of different denominations \u0026 an integer amount representing a total amount of money. Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.\n ๐Ÿฃ: 1๏ธโƒฃ Input: coins = [1,2,5], amount = 11 Output: 3 Explain: 11 = 5 + 5 + 1\n 2๏ธโƒฃ Input: coins = [2], amount = 3 Output: -1\n 3๏ธโƒฃ Input: coins = [1], amount = 0 Output: 0\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n^amount) ๐Ÿช: O(amount)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Memoization โฐ: O(n^amount) ๐Ÿช: O(n^amount) \n\n var coinChange = function (coins, amount) {\n let dp = new Array(amount + 1).fill(amount + 1);\n dp[0] = 0;\n for (let i = 1; i \u003c= amount; i++) {\n for (let j = 0; j \u003c coins.length; j++) {\n if (coins[j] \u003c= i) {\n dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);\n }\n }\n }\n return dp[amount] \u003e amount ? -1 : dp[amount];\n };\n## [Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) #300 ๐Ÿ“ˆ\n โ“: Given an integer array nums, return the length of the longest strictly increasing subsequence.\n ๐Ÿฃ: 1๏ธโƒฃ Input: nums = [10,9,2,5,3,7,101,18] Output: 4 Explain: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. 2๏ธโƒฃ Input: nums = [0,1,0,3,2,3] Output: 4 3๏ธโƒฃ Input: nums = [7,7,7,7,7,7,7] Output: 1\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(2^n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Memoization โฐ: O(n^2) ๐Ÿช: O(n^2) \n\n var lengthOfLIS = function (nums) {\n let dp = new Array(nums.length).fill(1);\n let max = 1;\n for (let i = 1; i \u003c nums.length; i++) {\n for (let j = 0; j \u003c i; j++) {\n if (nums[i] \u003e nums[j]) {\n dp[i] = Math.max(dp[i], dp[j] + 1);\n max = Math.max(max, dp[i]);\n }\n }\n }\n return max;\n };\n\n## [Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/) #1143 ๐Ÿ“ˆ\n โ“: Given two strings text1 \u0026 text2, return the length of their longest common subsequence. If there is no common subsequence, return 0. A subsequence of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, \"ace\" is a subsequence of \"abcde\" while \"aec\" is not). A common subsequence of two strings is a subsequence that is common to both strings.\n ๐Ÿฃ: 1๏ธโƒฃ Input: text1 = \"abcde\", text2 = \"ace\" Output: 3 Explain: The longest common subsequence is \"ace\" \u0026 its length is 3. 2๏ธโƒฃ Input: text1 = \"abc\", text2 = \"abc\" Output: 3 3๏ธโƒฃ Input: text1 = \"abc\", text2 = \"def\" Output: 0\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(2^n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Memoization โฐ: O(m*n) ๐Ÿช: O(m*n)\n\n var longestCommonSubsequence = function (text1, text2) {\n let dp = new Array(text1.length + 1)\n .fill(0)\n .map(() =\u003e new Array(text2.length + 1).fill(0));\n for (let i = 1; i \u003c= text1.length; i++) {\n for (let j = 1; j \u003c= text2.length; j++) {\n if (text1[i - 1] == text2[j - 1]) {\n dp[i][j] = dp[i - 1][j - 1] + 1;\n } else {\n dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);\n }\n }\n }\n return dp[text1.length][text2.length];\n };\n## [Word Break Problem](https://leetcode.com/problems/word-break/)๐Ÿ“ˆ\n โ“: Given a string s \u0026 a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words. Note that the same word in the dictionary may be reused multiple times in the segmentation.\n ๐Ÿฃ: 1๏ธโƒฃ Input: s = \"leetcode\", wordDict = [\"leet\",\"code\"] Output: true Explain: Return true because \"leetcode\" can be segmented as \"leet code\". 2๏ธโƒฃ Input: s = \"applepenapple\", wordDict = [\"apple\",\"pen\"] Output: true Explain: Return true because \"applepenapple\" can be segmented as \"apple pen apple\". Note that you are allowed to reuse a dictionary word. 3๏ธโƒฃ Input: s = \"cats\u0026og\", wordDict = [\"cats\",\"dog\",\"s\u0026\",\"\u0026\",\"cat\"] Output: false\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(2^n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Memoization โฐ: O(n^2) ๐Ÿช: O(n^2)\n\n var wordBreak = function (s, wordDict) {\n let dp = new Array(s.length + 1).fill(false);\n dp[0] = true;\n for (let i = 1; i \u003c= s.length; i++) {\n for (let j = 0; j \u003c i; j++) {\n if (dp[j] \u0026\u0026 wordDict.includes(s.substring(j, i))) {\n dp[i] = true;\n break;\n }\n }\n }\n return dp[s.length];\n };\n## [Combination Sum](https://leetcode.com/problems/combination-sum-iv/) #377 ๐Ÿ“ˆ\n โ“: Given an integer array with all positive numbers \u0026 no duplicates, find the number of possible combinations that add up to a positive integer target. \n ๐Ÿฃ: Input: nums = [1, 2, 3] target = 4 Output: 7 Explain: The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 3) (2, 1, 1) (2, 2) (3, 1) Note that different sequences are counted as different combinations. Therefore the output is 7.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(2^n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Memoization โฐ: O(n*target) ๐Ÿช: O(n*target)\n\n var combinationSum4 = function (nums, target) {\n let dp = new Array(target + 1).fill(0);\n dp[0] = 1;\n for (let i = 1; i \u003c= target; i++) {\n for (let j = 0; j \u003c nums.length; j++) {\n if (nums[j] \u003c= i) {\n dp[i] += dp[i - nums[j]];\n }\n }\n }\n return dp[target];\n };\n## [House Robber](https://leetcode.com/problems/house-robber/) #198\n โ“: You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constrastopping you from robbing each of them is that adjacent houses have security system connected \u0026 it will automatically contact the police if two adjacent houses were broken into on the same night. Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.\n ๐Ÿฃ: 1๏ธโƒฃ Input: nums = [1,2,3,1] Output: 4 Explain: Rob house 1 (money = 1) \u0026 then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4. 2๏ธโƒฃ Input: nums = [2,7,9,3,1] Output: 12 Explain: Rob house 1 (money = 2), rob house 3 (money = 9) \u0026 rob house 5 (money = 1). Total amount you can rob = 2 + 9 + 1 = 12.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(2^n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Memoization โฐ: O(n) ๐Ÿช: O(n)\n\n var rob = function (nums) {\n let dp = new Array(nums.length + 1).fill(0);\n dp[0] = 0;\n dp[1] = nums[0];\n for (let i = 2; i \u003c= nums.length; i++) {\n dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i - 1]);\n }\n return dp[nums.length];\n };\n## [House Robber II](https://leetcode.com/problems/house-robber-ii/) #213\n\n โ“: You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected \u0026 it will automatically contact the police if two adjacent houses were broken into on the same night. Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.\n ๐Ÿฃ: 1๏ธโƒฃ Input: nums = [2,3,2] Output: 3 Explain: You cannot rob house 1 (money = 2) \u0026 then rob house 3 (money = 2), because they are adjacent houses. 2๏ธโƒฃ Input: nums = [1,2,3,1] Output: 4 Explain: Rob house 1 (money = 1) \u0026 then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4. 3๏ธโƒฃ Input: nums = [0] Output: 0\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(2^n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Memoization โฐ: O(n) ๐Ÿช: O(n)\n\n var rob = function (nums) {\n if (nums.length == 1) return nums[0];\n return Math.max(\n robRange(nums, 0, nums.length - 2),\n robRange(nums, 1, nums.length - 1)\n );\n };\n var robRange = function (nums, start, end) {\n let dp = new Array(nums.length + 1).fill(0);\n dp[start] = nums[start];\n dp[start + 1] = Math.max(nums[start], nums[start + 1]);\n for (let i = start + 2; i \u003c= end; i++) {\n dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);\n }\n return dp[end];\n };\n## [Decode Ways](https://leetcode.com/problems/decode-ways/) #91 ๐Ÿ“ˆ\n โ“: A message containing letters from A-Z is being encoded to numbers using the following mapping: 'A' -\u003e 1 'B' -\u003e 2 ... 'Z' -\u003e 26 Given a non-empty string s containing only digits, determine the total number of ways to decode it. The answer is guaranteed to fit in a 32-bit integer.\n ๐Ÿฃ: 1๏ธโƒฃ Input: s = \"12\" Output: 2 Explain: It could be decoded as \"AB\" (1 2) or \"L\" (12). 2๏ธโƒฃ Input: s = \"226\" Output: 3 Explain: It could be decoded as \"BZ\" (2 26), \"VF\" (22 6), or \"BBF\" (2 2 6). 3๏ธโƒฃ Input: s = \"0\" Output: 0 Explain: There is no character that is mapped to a number starting with 0. The only valid mappings with 0 are 'J' -\u003e \"10\" \u0026 'T' -\u003e \"20\", neither of which start with 0. Hence, there are no valid ways to decode this since all digits need to be mapped. #4 Input: s = \"1\" Output: 1\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(2^n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Memoization โฐ: O(n) ๐Ÿช: O(n)\n\n var numDecodings = function (s) {\n if (s[0] == \"0\") return 0;\n let dp = new Array(s.length + 1).fill(0);\n dp[0] = 1;\n dp[1] = 1;\n for (let i = 2; i \u003c= s.length; i++) {\n let first = parseInt(s.substring(i - 1, i));\n let second = parseInt(s.substring(i - 2, i));\n if (first \u003e= 1 \u0026\u0026 first \u003c= 9) {\n dp[i] += dp[i - 1];\n }\n if (second \u003e= 10 \u0026\u0026 second \u003c= 26) {\n dp[i] += dp[i - 2];\n }\n }\n return dp[s.length];\n };\n## [Unique Paths](https://leetcode.com/problems/unique-paths/) #62 ๐Ÿ“ˆ\n\n โ“: A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any poin time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). How many possible unique paths are there?\n ๐Ÿฃ: 1๏ธโƒฃ Input: m = 3, n = 7 Output: 28 2๏ธโƒฃ Input: m = 3, n = 2 Output: 3 Explain: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner: 1. Right -\u003e Down -\u003e Down 2. Down -\u003e Down -\u003e Right 3. Down -\u003e Right -\u003e Down 3๏ธโƒฃ Input: m = 7, n = 3 Output: 28 #4 Input: m = 3, n = 3 Output: 6\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(2^n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Memoization โฐ: O(n) ๐Ÿช: O(n)\n\n var uniquePaths = function (m, n) {\n let dp = new Array(m).fill(0).map(() =\u003e new Array(n).fill(0));\n for (let i = 0; i \u003c m; i++) {\n dp[i][0] = 1;\n }\n for (let j = 0; j \u003c n; j++) {\n dp[0][j] = 1;\n }\n for (let i = 1; i \u003c m; i++) {\n for (let j = 1; j \u003c n; j++) {\n dp[i][j] = dp[i - 1][j] + dp[i][j - 1];\n }\n }\n return dp[m - 1][n - 1];\n };\n## [Jump Game](https://leetcode.com/problems/jump-game/) #55 ๐Ÿ“ˆ\n\n โ“: Given an array of non-negative integers nums, you are initially positioned at the first index of the array. Each element in the array represents your maximum jump length at that position. Determine if you are able to reach the last index.\n ๐Ÿฃ: 1๏ธโƒฃ Input: nums = [2,3,1,1,4] Output: true Explain: Jump 1 step from index 0 to 1, then 3 steps to the last index. 2๏ธโƒฃ Input: nums = [3,2,1,0,4] Output: false Explain: You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible to reach the last index.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(2^n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming Tabulation โฐ: O(n) ๐Ÿช: O(n)\n\n var canJump = function (nums) {\n let dp = new Array(nums.length).fill(false);\n dp[0] = true;\n for (let i = 1; i \u003c nums.length; i++) {\n for (let j = 0; j \u003c i; j++) {\n if (dp[j] \u0026\u0026 j + nums[j] \u003e= i) {\n dp[i] = true;\n break;\n }\n }\n }\n return dp[nums.length - 1];\n };\n\n---\n## Graph\n\n## [Clone Graph](https://leetcode.com/problems/clone-graph/) #133 ๐ŸŒณ\n โ“: Given a reference of a node in a connected undirected graph. Return a deep copy (clone) of the graph. Each node in the graph contains a val (int) \u0026 a list (List[Node]) of its neighbors.\n ๐Ÿฃ: 1๏ธโƒฃ Input: adjList = [[2,4],[1,3],[2,4],[1,3]] Output: [[2,4],[1,3],[2,4],[1,3]] Explain: There are 4 nodes in the graph. 2๏ธโƒฃ Input: adjList = [[]] Output: [[]] 3๏ธโƒฃ Input: adjList = [] Output: [] #4 Input: adjList = [[2],[1]] Output: [[2],[1]]\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐ŸŒณ BFS โฐ: O(n) ๐Ÿช: O(n)\n\n var cloneGraph = function (node) {\n if (!node) return node;\n let map = new Map();\n let clone = new Node(node.val, []);\n map.set(node, clone);\n let queue = [node];\n while (queue.length) {\n let n = queue.shift();\n for (let neighbor of n.neighbors) {\n if (!map.has(neighbor)) {\n map.set(neighbor, new Node(neighbor.val, []));\n queue.push(neighbor);\n }\n map.get(n).neighbors.push(map.get(neighbor));\n }\n }\n return clone;\n };\n\n## [Course Schedule](https://leetcode.com/problems/course-schedule/) #207 ๐Ÿ“…\n โ“: There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai. For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1. Return true if you can finish all courses. Otherwise, return false.\n ๐Ÿฃ: 1๏ธโƒฃ Input: numCourses = 2, prerequisites = [[1,0]] Output: true Explain: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible. 2๏ธโƒฃ Input: numCourses = 2, prerequisites = [[1,0],[0,1]] Output: false Explain: There are a total of 2 courses to take. To take course 1 you should have finished course 0, \u0026 to take course 0 you should also have finished course 1. So it is impossible.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿ“… Topological Sort โฐ: O(V+E) ๐Ÿช: O(V+E) \n\n var canFinish = function (numCourses, prerequisites) {\n let graph = new Map();\n let indegree = new Array(numCourses).fill(0);\n for (let [course, pre] of prerequisites) {\n if (!graph.has(pre)) graph.set(pre, []);\n graph.get(pre).push(course);\n indegree[course]++;\n }\n let queue = [];\n for (let i = 0; i \u003c indegree.length; i++) {\n if (indegree[i] == 0) queue.push(i);\n }\n let count = 0;\n while (queue.length) {\n let course = queue.shift();\n count++;\n if (graph.has(course)) {\n for (let next of graph.get(course)) {\n indegree[next]--;\n if (indegree[next] == 0) queue.push(next);\n }\n }\n }\n return count == numCourses;\n };\n\n## [Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/) #417 ๐ŸŒณ \n โ“: Given an m x n matrix of non-negative integers representing the height of each unit cell in a continent, the \"Pacific ocean\" touches the left \u0026 top edges of the matrix \u0026 the \"Atlantic ocean\" touches the right \u0026 bottom edges. Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower. Find the list of grid coordinates where water can flow to both the Pacific \u0026 Atlantic ocean.\n ๐Ÿฃ: 1๏ธโƒฃ Input: heights = [[1,2,2,3,5],[3,2,3,4,4],[2,4,5,3,1],[6,7,1,4,5],[5,1,1,2,4]] Output: [[0,4],[1,3],[1,4],[2,2],[3,0],[3,1],[4,0]] Explain: Pacific ~ ~ ~ ~ ~ 1 2 2 3 (5) ~ 3 2 3 (4) (4) ~ 2 4 (5) 3 1 ~ (6) (7) 1 4 5 ~ (5) 1 1 2 4 ~ ~ ~ ~ ~ Atlantic 2๏ธโƒฃ Input: heights = [[2,1],[1,2]] Output: [[0,0],[0,1],[1,0],[1,1]]\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐ŸŒณ BFS โฐ: O(n) ๐Ÿช: O(n)\n\n var pacificAtlantic = function (heights) {\n let m = heights.length;\n let n = heights[0].length;\n let pacific = new Array(m).fill(0).map(() =\u003e new Array(n).fill(false));\n let atlantic = new Array(m).fill(0).map(() =\u003e new Array(n).fill(false));\n let queue = [];\n for (let i = 0; i \u003c m; i++) {\n queue.push([i, 0]);\n pacific[i][0] = true;\n }\n for (let i = 0; i \u003c n; i++) {\n queue.push([0, i]);\n pacific[0][i] = true;\n }\n ๐ŸŒณ BFS(heights, pacific, queue);\n queue = [];\n for (let i = 0; i \u003c m; i++) {\n queue.push([i, n - 1]);\n atlantic[i][n - 1] = true;\n }\n for (let i = 0; i \u003c n; i++) {\n queue.push([m - 1, i]);\n atlantic[m - 1][i] = true;\n }\n ๐ŸŒณ BFS(heights, atlantic, queue);\n let res = [];\n for (let i = 0; i \u003c m; i++) {\n for (let j = 0; j \u003c n; j++) {\n if (pacific[i][j] \u0026\u0026 atlantic[i][j]) res.push([i, j]);\n }\n }\n return res;\n };\n\n function ๐ŸŒณ BFS(heights, ocean, queue) {\n let dirs = [[0, 1], [0, -1], [1, 0], [-1, 0]];\n while (queue.length) {\n let [i, j] = queue.shift();\n for (let dir of dirs) {\n let x = i + dir[0];\n let y = j + dir[1];\n if (\n x \u003e= 0 \u0026\u0026\n x \u003c heights.length \u0026\u0026\n y \u003e= 0 \u0026\u0026\n y \u003c heights[0].length \u0026\u0026\n !ocean[x][y]\n ) {\n if (heights[x][y] \u003e= heights[i][j]) {\n ocean[x][y] = true;\n queue.push([x, y]);\n }\n }\n }\n }\n }\n\n\n \n## [Number of Islands](https://leetcode.com/problems/number-of-islands/) #200 ๐ŸŒณ\n Pattern Used: ๐ŸŒณ BFS Pattern ๐ŸŒฒ DFS Pattern Union Find Pattern\n โ“: Given an m x n 2d grid map of '1's (l\u0026) \u0026 '0's (water), return the number of islands. An isl\u0026 is surrounded by water \u0026 is formed by connecting adjacent l\u0026s horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.\n ๐Ÿฃ: 1๏ธโƒฃ Input: grid = [ [\"1\",\"1\",\"1\",\"1\",\"0\"], [\"1\",\"1\",\"0\",\"1\",\"0\"], [\"1\",\"1\",\"0\",\"0\",\"0\"], [\"0\",\"0\",\"0\",\"0\",\"0\"] ] Output: 1 2๏ธโƒฃ Input: grid = [ [\"1\",\"1\",\"0\",\"0\",\"0\"], [\"1\",\"1\",\"0\",\"0\",\"0\"], [\"0\",\"0\",\"1\",\"0\",\"0\"], [\"0\",\"0\",\"0\",\"1\",\"1\"] ] Output: 3\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐ŸŒณ BFS โฐ: O(mn) ๐Ÿช: O(mn)\n\n var numIslands = function (grid) {\n let m = grid.length;\n let n = grid[0].length;\n let count = 0;\n let dirs = [[0, 1], [0, -1], [1, 0], [-1, 0]];\n for (let i = 0; i \u003c m; i++) {\n for (let j = 0; j \u003c n; j++) {\n if (grid[i][j] == \"1\") {\n count++;\n grid[i][j] = \"0\";\n let queue = [[i, j]];\n while (queue.length) {\n let [x, y] = queue.shift();\n for (let dir of dirs) {\n let a = x + dir[0];\n let b = y + dir[1];\n if (\n a \u003e= 0 \u0026\u0026\n a \u003c m \u0026\u0026\n b \u003e= 0 \u0026\u0026\n b \u003c n \u0026\u0026\n grid[a][b] == \"1\"\n ) {\n grid[a][b] = \"0\";\n queue.push([a, b]);\n }\n }\n }\n }\n }\n }\n return count;\n };\n## [Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/) #128 โ™พ \n โ“: Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence. You must write an algorithm that runs in O(n) time.\n ๐Ÿฃ: 1๏ธโƒฃ Input: nums = [100,4,200,1,3,2] Output: 4 Explain: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4. 2๏ธโƒฃ Input: nums = [0,3,7,2,5,8,4,6,0,1] Output: 9\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: โ™พ Union Find โฐ: O(n) ๐Ÿช: O(n)\n\n var longestConsecutive = function (nums) {\n let set = new Set(nums);\n let max = 0;\n for (let num of set) {\n if (!set.has(num - 1)) {\n let curr = num;\n let count = 1;\n while (set.has(curr + 1)) {\n curr++;\n count++;\n }\n max = Math.max(max, count);\n }\n }\n return max;\n };\n\n## Interval\n\n## [Insert Interval](https://leetcode.com/problems/insert-interval/) #57 ๐Ÿš—๐Ÿš™ \n โ“: Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary). You may assume that the intervals were initially sorted according to their start times.\n ๐Ÿฃ: 1๏ธโƒฃ Input: intervals = [[1,3],[6,9]], newInterval = [2,5] Output: [[1,5],[6,9]] 2๏ธโƒฃ Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8] Output: [[1,2],[3,10],[12,16]] Explain: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿš—๐Ÿš™ Merge Intervals โฐ: O(n) ๐Ÿช: O(n)\n \n var insert = function (intervals, newInterval) {\n let res = [];\n let i = 0;\n while (i \u003c intervals.length \u0026\u0026 intervals[i][1] \u003c newInterval[0]) {\n res.push(intervals[i]);\n i++;\n }\n while (i \u003c intervals.length \u0026\u0026 intervals[i][0] \u003c= newInterval[1]) {\n newInterval[0] = Math.min(newInterval[0], intervals[i][0]);\n newInterval[1] = Math.max(newInterval[1], intervals[i][1]);\n i++;\n }\n res.push(newInterval);\n while (i \u003c intervals.length) {\n res.push(intervals[i]);\n i++;\n }\n return res;\n };\n## [Merge Intervals](https://leetcode.com/problems/merge-intervals/) #56 ๐Ÿš—๐Ÿš™ ๐Ÿ’ฐ\n โ“: Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, \u0026 return an array of the non-overlapping intervals that cover all the intervals in the input.\n ๐Ÿฃ: 1๏ธโƒฃ Input: intervals = [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]] Explain: Since intervals [1,3] \u0026 [2,6] overlaps, merge them into [1,6]. 2๏ธโƒฃ Input: intervals = [[1,4],[4,5]] Output: [[1,5]] Explain: Intervals [1,4] \u0026 [4,5] are considered overlapping.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿš—๐Ÿš™ Merge Intervals โฐ: O(n) ๐Ÿช: O(n)\n \n var merge = function (intervals) {\n if (intervals.length \u003c= 1) return intervals;\n intervals.sort((a, b) =\u003e a[0] - b[0]);\n let res = [intervals[0]];\n for (let i = 1; i \u003c intervals.length; i++) {\n let curr = intervals[i];\n let prev = res[res.length - 1];\n if (curr[0] \u003c= prev[1]) {\n prev[1] = Math.max(prev[1], curr[1]);\n } else {\n res.push(curr);\n }\n }\n return res;\n };\n\n## [Non-overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/) #435 ๐Ÿš—๐Ÿš™\n โ“: Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.\n ๐Ÿฃ: 1๏ธโƒฃ Input: intervals = [[1,2],[2,3],[3,4],[1,3]] Output: 1 Explain: [1,3] can be removed \u0026 the rest of the intervals are non-overlapping. 2๏ธโƒฃ Input: intervals = [[1,2],[1,2],[1,2]] Output: 2 Explain: You need to remove two [1,2] to make the rest of the intervals non-overlapping. 3๏ธโƒฃ Input: intervals = [[1,2],[2,3]] Output: 0 Explain: You don't need to remove any of the intervals since they're already non-overlapping.\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿš—๐Ÿš™ Merge Intervals โฐ: O(nlogn) ๐Ÿช: O(1)\n\n var eraseOverlapIntervals = function (intervals) {\n if (intervals.length \u003c= 1) return 0;\n intervals.sort((a, b) =\u003e a[1] - b[1]);\n let count = 0;\n let prev = intervals[0];\n for (let i = 1; i \u003c intervals.length; i++) {\n let curr = intervals[i];\n if (curr[0] \u003c prev[1]) {\n count++;\n } else {\n prev = curr;\n }\n }\n return count;\n };\n---\n\n## Linked List\n\n## [Reverse a Linked List](https://leetcode.com/problems/reverse-linked-list/) #206 ๐Ÿฐ\u0026๐Ÿข \n โ“: Given the head of a singly linked list, reverse the list, \u0026 return the reversed list.\n ๐Ÿฃ: 1๏ธโƒฃ Input: head = [1,2,3,4,5] Output: [5,4,3,2,1] 2๏ธโƒฃ Input: head = [1,2] Output: [2,1] 3๏ธโƒฃ Input: head = [] Output: []\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: ๐Ÿ”จ Brute Force โฐ:\n ๐Ÿ‡ Solution: ๐Ÿฐ\u0026๐Ÿข Fast \u0026 Slow Pointers โฐ: O(n) ๐Ÿช: O(1)\n\n var reverseList = function (head) {\n let prev = null;\n let curr = head;\n while (curr) {\n let next = curr.next;\n curr.next = prev;\n prev = curr;\n curr = next;\n }\n return prev;\n };\n which algorithm from ./algorithms.md is used in this solution?\n the \n## [Detect Cycle in a Linked List](https://leetcode.com/problems/linked-list-cycle/) #141 ๐Ÿฐ\u0026๐Ÿข \n โ“: Given head, the head of a linked list, determine if the linked list has a cycle in it. There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter. Return true if there is a cycle in the linked list. Otherwise, return false.\n ๐Ÿฃ: 1๏ธโƒฃ Input: head = [3,2,0,-4], pos = 1 Output: true Explain: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed). 2๏ธโƒฃ Input: head = [1,2], pos = 0 Output: true Explain: There is a cycle in the linked list, where the tail connects to the 0th node. 3๏ธโƒฃ Input: head = [1], pos = -1 Output: false Explain: There is no cycle in the linked list.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿฐ\u0026๐Ÿข Fast \u0026 Slow Pointers โฐ: O(n) ๐Ÿช: O(1)\n\n var hasCycle = function (head) {\n let slow = head;\n let fast = head;\n while (fast \u0026\u0026 fast.next) {\n slow = slow.next;\n fast = fast.next.next;\n if (slow === fast) return true;\n }\n return false;\n };\n## [Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/) #21 ๐Ÿฐ\u0026๐Ÿข \n โ“: Merge two sorted linked lists \u0026 return it as a sorted list. The list should be made by splicing together the nodes of the first two lists.\n ๐Ÿฃ: 1๏ธโƒฃ Input: l1 = [1,2,4], l2 = [1,3,4] Output: [1,1,2,3,4,4] 2๏ธโƒฃ Input: l1 = [], l2 = [] Output: [] 3๏ธโƒฃ Input: l1 = [], l2 = [0] Output: [0]\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿฐ\u0026๐Ÿข Fast \u0026 Slow Pointers โฐ: O(n) ๐Ÿช: O(1)\n\n var mergeTwoLists = function (l1, l2) {\n let dummy = new ListNode();\n let curr = dummy;\n while (l1 \u0026\u0026 l2) {\n if (l1.val \u003c l2.val) {\n curr.next = l1;\n l1 = l1.next;\n } else {\n curr.next = l2;\n l2 = l2.next;\n }\n curr = curr.next;\n }\n if (l1) curr.next = l1;\n if (l2) curr.next = l2;\n return dummy.next;\n };\n ## [Merge K Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/) #23\n\n Pattern Used: K MERGE \n โ“: You are given an array of k linked-lists lists, each linked-list is sorted in ascending order. Merge all the linked-lists into one sorted linked-list \u0026 return it.\n ๐Ÿฃ: 1๏ธโƒฃ Input: lists = [[1,4,5],[1,3,4],[2,6]] Output: [1,1,2,3,4,4,5,6] 2๏ธโƒฃ Input: lists = [] Output: [] 3๏ธโƒฃ Input: lists = [[]] Output: []\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: Heap โฐ: O(n) ๐Ÿช: O(n)\n\n import ListNode from \"DataStructures/LinkedList/ListNode.js\";\n import MinHeap from \"DataStructures/Heaps/MinHeap.js\";\n\n const mergeKLists = function (lists) {\n if (lists.length === 0) return null;\n let dummy = new ListNode();\n let curr = dummy;\n let minHeap = new MinHeap();\n for (let i = 0; i \u003c lists.length; i++) {\n if (lists[i]) minHeap.insert(lists[i]);\n }\n while (minHeap.size() \u003e 0) {\n let node = minHeap.remove();\n curr.next = node;\n curr = curr.next;\n if (node.next) minHeap.insert(node.next);\n }\n return dummy.next;\n };\n## [Remove Nth Node From End Of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) #19 ๐Ÿฐ\u0026๐Ÿข\n โ“: Given the head of a linked list, remove the nth node from the end of the list \u0026 return its head.\n ๐Ÿฃ: 1๏ธโƒฃ Input: head = [1,2,3,4,5], n = 2 Output: [1,2,3,5] 2๏ธโƒฃ Input: head = [1], n = 1 Output: [] 3๏ธโƒฃ Input: head = [1,2], n = 1 Output: [1]\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿฐ\u0026๐Ÿข Fast \u0026 Slow Pointers Fast \u0026 Slow Pointers โฐ: O(n) ๐Ÿช: O(1)\n\n var removeNthFromEnd = function (head, n) {\n let dummy = new ListNode();\n dummy.next = head;\n let slow = dummy;\n let fast = dummy;\n for (let i = 0; i \u003c n; i++) {\n fast = fast.next;\n }\n while (fast.next) {\n slow = slow.next;\n fast = fast.next;\n }\n slow.next = slow.next.next;\n return dummy.next;\n };\n## [Reorder List](https://leetcode.com/problems/reorder-list/) #143 ๐Ÿฐ\u0026๐Ÿข\n โ“: Given the head of a singly linked list, reorder the list to be: head -\u003e node 2 -\u003e node 3 -\u003e node 4 -\u003e ... -\u003e node n -\u003e null. You may not modify the values in the list's nodes. Only nodes themselves may be changed.\n ๐Ÿฃ: 1๏ธโƒฃ Input: head = [1,2,3,4] Output: [1,4,2,3] 2๏ธโƒฃ Input: head = [1,2,3,4,5] Output: [1,5,2,4,3]\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐Ÿฐ\u0026๐Ÿข Fast \u0026 Slow Pointers โฐ: O(n) ๐Ÿช: O(1)\n\n var reorderList = function (head) {\n if (!head) return null;\n let slow = head;\n let fast = head;\n while (fast \u0026\u0026 fast.next) {\n slow = slow.next;\n fast = fast.next.next;\n }\n let prev = null;\n let curr = slow;\n while (curr) {\n let next = curr.next;\n curr.next = prev;\n prev = curr;\n curr = next;\n }\n let first = head;\n let second = prev;\n while (second.next) {\n let temp = first.next;\n first.next = second;\n first = temp;\n temp = second.next;\n second.next = first;\n second = temp;\n }\n };\n---\n\n## Matrix\n\n## [Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/) #73 ๐Ÿ“\n โ“: Given an m x n matrix. If an element is 0, set its entire row \u0026 column to 0. Do it in-place.\n ๐Ÿฃ: 1๏ธโƒฃ Input: matrix = [[1,1,1],[1,0,1],[1,1,1]] Output: [[1,0,1],[0,0,0],[1,0,1]] 2๏ธโƒฃ Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]] Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(mn) ๐Ÿช: O(m+n)\n ๐Ÿ‡ Solution: ๐Ÿ“ Matrix Pattern โฐ: O(mn) ๐Ÿช: O(1)\n\n var setZeroes = function (matrix) {\n let rows = new Set();\n let cols = new Set();\n for (let i = 0; i \u003c matrix.length; i++) {\n for (let j = 0; j \u003c matrix[0].length; j++) {\n if (matrix[i][j] === 0) {\n rows.add(i);\n cols.add(j);\n }\n }\n }\n for (let i = 0; i \u003c matrix.length; i++) {\n for (let j = 0; j \u003c matrix[0].length; j++) {\n if (rows.has(i) || cols.has(j)) {\n matrix[i][j] = 0;\n }\n }\n }\n };\n\n## [Spiral Matrix](https://leetcode.com/problems/spiral-matrix/) #54 ๐Ÿ“\n Pattern Used: ๐Ÿ“ Matrix Pattern\n โ“: Given an m x n matrix, return all elements of the matrix in spiral order.\n ๐Ÿฃ: 1๏ธโƒฃ Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,2,3,6,9,8,7,4,5] 2๏ธโƒฃ Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]] Output: [1,2,3,4,8,12,11,10,9,5,6,7]\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(mn) ๐Ÿช: O(mn)\n ๐Ÿ‡ Solution: ๐Ÿ“ Matrix Pattern โฐ: O(mn) ๐Ÿช: O(mn)\n\n var spiralOrder = function (matrix) {\n let result = [];\n let top = 0;\n let bottom = matrix.length - 1;\n let left = 0;\n let right = matrix[0].length - 1;\n while (top \u003c= bottom \u0026\u0026 left \u003c= right) {\n for (let i = left; i \u003c= right; i++) {\n result.push(matrix[top][i]);\n }\n top++;\n for (let i = top; i \u003c= bottom; i++) {\n result.push(matrix[i][right]);\n }\n right--;\n if (top \u003c= bottom) {\n for (let i = right; i \u003e= left; i--) {\n result.push(matrix[bottom][i]);\n }\n bottom--;\n }\n if (left \u003c= right) {\n for (let i = bottom; i \u003e= top; i--) {\n result.push(matrix[i][left]);\n }\n left++;\n }\n }\n return result;\n };\n## [Rotate Image](https://leetcode.com/problems/rotate-image/) #48 ๐Ÿ“\n Pattern Used: ๐Ÿ“ Matrix Pattern\n โ“: You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).\n ๐Ÿฃ: 1๏ธโƒฃ Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [[7,4,1],[8,5,2],[9,6,3]] 2๏ธโƒฃ Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]] Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: ๐Ÿ”จ Brute Force โฐ:\n ๐Ÿ‡ Solution: MatrixPattern โฐ: O(n) ๐Ÿช: O(1)\n\n var rotate = function (matrix) {\n let n = matrix.length;\n for (let i = 0; i \u003c n / 2; i++) {\n for (let j = i; j \u003c n - i - 1; j++) {\n let temp = matrix[i][j];\n matrix[i][j] = matrix[n - j - 1][i];\n matrix[n - j - 1][i] = matrix[n - i - 1][n - j - 1];\n matrix[n - i - 1][n - j - 1] = matrix[j][n - i - 1];\n matrix[j][n - i - 1] = temp;\n }\n }\n };\n## [Word Search](https://leetcode.com/problems/word-search/) #79 ๐Ÿ“\n Pattern Used: Backtracking ๐Ÿ“ Matrix Pattern\n โ“: Given an m x n grid of characters board \u0026 a string word, return true if word exists in the grid.\n ๐Ÿฃ: 1๏ธโƒฃ Input: board = [[\"A\",\"B\",\"C\",\"E\"],[\"S\",\"F\",\"C\",\"S\"],[\"A\",\"D\",\"E\",\"E\"]], word = \"ABCCED\" Output: true 2๏ธโƒฃ Input: board = [[\"A\",\"B\",\"C\",\"E\"],[\"S\",\"F\",\"C\",\"S\"],[\"A\",\"D\",\"E\",\"E\"]], word = \"SEE\" Output: true 3๏ธโƒฃ Input: board = [[\"A\",\"B\",\"C\",\"E\"],[\"S\",\"F\",\"C\",\"S\"],[\"A\",\"D\",\"E\",\"E\"]], word = \"ABCB\" Output: false\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ O(mn * 4^L) ๐Ÿช: O(mn)\n ๐Ÿ‡ Solution: MatrixPattern โฐ: O(n) ๐Ÿช: O(n)\n\n var exist = function (board, word) {\n let m = board.length;\n let n = board[0].length;\n let visited = new Array(m);\n for (let i = 0; i \u003c m; i++) {\n visited[i] = new Array(n).fill(false);\n }\n let ๐ŸŒฒ DFS = (i, j, k) =\u003e {\n if (i \u003c 0 || i \u003e= m || j \u003c 0 || j \u003e= n || visited[i][j] || board[i][j] !== word[k]) {\n return false;\n }\n if (k === word.length - 1) {\n return true;\n }\n visited[i][j] = true;\n let result =\n ๐ŸŒฒ DFS(i + 1, j, k + 1) ||\n ๐ŸŒฒ DFS(i - 1, j, k + 1) ||\n ๐ŸŒฒ DFS(i, j + 1, k + 1) ||\n ๐ŸŒฒ DFS(i, j - 1, k + 1);\n visited[i][j] = false;\n return result;\n };\n for (let i = 0; i \u003c m; i++) {\n for (let j = 0; j \u003c n; j++) {\n if (๐ŸŒฒ DFS(i, j, 0)) {\n return true;\n }\n }\n }\n return false;\n };\n\n---\n\n## String\n\n\n## [Valid Anagram](https://leetcode.com/problems/valid-anagram/) #242 \n โ“: Given two strings s \u0026 t , write a function to determine if t is an anagram of s.\n ๐Ÿฃ: 1๏ธโƒฃ Input: s = \"anagram\", t = \"nagaram\" Output: true 2๏ธโƒฃ Input: s = \"rat\", t = \"car\" Output: false\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(nlogn) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: Hash Table โฐ: O(n) ๐Ÿช: O(n)\n \n var isAnagram = function (s, t) {\n if (s.length !== t.length) {\n return false;\n }\n let map = new Map();\n for (let i = 0; i \u003c s.length; i++) {\n map.set(s[i], (map.get(s[i]) || 0) + 1);\n }\n for (let i = 0; i \u003c t.length; i++) {\n if (!map.has(t[i])) {\n return false;\n }\n map.set(t[i], map.get(t[i]) - 1);\n if (map.get(t[i]) \u003c 0) {\n return false;\n }\n }\n return true;\n };\n## [Group Anagrams](https://leetcode.com/problems/group-anagrams/) #49\n โ“: Given an array of strings strs, group the anagrams together. You can return the answer in any order.\n An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n ๐Ÿฃ: 1๏ธโƒฃ Input: strs = [\"eat\",\"tea\",\"tan\",\"ate\",\"nat\",\"bat\"] Output: [[\"bat\"],[\"nat\",\"tan\"],[\"ate\",\"eat\",\"tea\"]] 2๏ธโƒฃ Input: strs = [\"\"] Output: [[\"\"]] 3๏ธโƒฃ Input: strs = [\"a\"] Output: [[\"a\"]]\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n^2)\n ๐Ÿ‡ Solution: Hash Table โฐ: O(n) ๐Ÿช: O(n)\n\n var groupAnagrams = function (strs) {\n let map = new Map();\n for (let i = 0; i \u003c strs.length; i++) {\n let sorted = strs[i].split(\"\").sort().join(\"\");\n if (!map.has(sorted)) {\n map.set(sorted, []);\n }\n map.get(sorted).push(strs[i]);\n }\n return [...map.values()];\n };\n## [Valid Parentheses](https://leetcode.com/problems/valid-parentheses/) #20\n Pattern Used: Stack\n โ“: Given a string s containing just the characters '(', ')', '{', '}', '[' \u0026 ']', determine if the input string is valid.\n An input string is valid if:\n Open brackets must be closed by the same type of brackets.\n Open brackets must be closed in the correct order.\n ๐Ÿฃ: 1๏ธโƒฃ Input: s = \"()\" Output: true 2๏ธโƒฃ Input: s = \"()[]{}\" Output: true 3๏ธโƒฃ Input: s = \"(]\" Output: false #4 Input: s = \"([)]\" Output: false #5 Input: s = \"{[]}\" Output: true\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n^2) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: Stack โฐ: O(n) ๐Ÿช: O(n)\n\n var isValid = function (s) {\n let stack = [];\n let map = new Map();\n map.set(\"(\", \")\");\n map.set(\"{\", \"}\");\n map.set(\"[\", \"]\");\n for (let i = 0; i \u003c s.length; i++) {\n if (map.has(s[i])) {\n stack.push(s[i]);\n } else {\n if (map.get(stack.pop()) !== s[i]) {\n return false;\n }\n }\n }\n return stack.length === 0;\n };\n## [Valid Palindrome](https://leetcode.com/problems/valid-palindrome/) #125 ๐Ÿ‘ฏ\n โ“: Given a string s, determine if it is a palindrome, considering only alphanumeric characters \u0026 ignoring cases.\n ๐Ÿฃ: 1๏ธโƒฃ Input: s = \"A man, a plan, a canal: Panama\" Output: true Explain: \"amanaplanacanalpanama\" is a palindrome.\n \n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: \n ๐Ÿ‡ Solution: ๐Ÿ‘ฏ Two Pointers โฐ: \n\n var isPalindrome = function(s) {\n let cursor1 = 0;\n let cursor2 = s.length - 1;\n\n while (cursor1 \u003c cursor2) {\n if (!(/^[a-zA-Z0-9]*$/.test(s[cursor1]))) {\n cursor1++;\n continue;\n }\n if (!(/^[a-zA-Z0-9]*$/.test(s[cursor2]))) {\n cursor2--;\n continue;\n }\n if (s[cursor1].toLowerCase() === s[cursor2].toLowerCase()) {\n cursor1++;\n cursor2--;\n continue;\n }\n return false;\n }\n\n return true;\n };\n \n }\n## [Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/) #5 ๐Ÿ“ˆ\n โ“: Given a string s, return the longest palindromic substring in s.\n ๐Ÿฃ: 1๏ธโƒฃ Input: s = \"babad\" Output: \"bab\" Note: \"aba\" is also a valid answer. 2๏ธโƒฃ Input: s = \"cbbd\" Output: \"bb\" 3๏ธโƒฃ Input: s = \"a\" Output: \"a\" #4 Input: s = \"ac\" Output: \"a\"\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n^3) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming โฐ: O(n^2) ๐Ÿช: O(n^2)\n\n var longestPalindrome = function (s) {\n let result = \"\";\n for (let i = 0; i \u003c s.length; i++) {\n let odd = exp\u0026(s, i, i);\n let even = exp\u0026(s, i, i + 1);\n let max = odd.length \u003e even.length ? odd : even;\n if (max.length \u003e result.length) {\n result = max;\n }\n }\n return result;\n };\n function exp\u0026(s, left, right) {\n while (left \u003e= 0 \u0026\u0026 right \u003c s.length \u0026\u0026 s[left] === s[right]) {\n left--;\n right++;\n }\n return s.substring(left + 1, right);\n }\n## [Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/) #647 ๐Ÿ“ˆ \n โ“: Given a string s, return the number of palindromic substrings in it.\n A string is a palindrome when it reads the same backward as forward.\n A substring is a contiguous sequence of characters within the string.\n ๐Ÿฃ: 1๏ธโƒฃ Input: s = \"abc\" Output: 3 Explain: Three palindromic strings: \"a\", \"b\", \"c\". 2๏ธโƒฃ Input: s = \"aaa\" Output: 6 Explain: Six palindromic strings: \"a\", \"a\", \"a\", \"aa\", \"aa\", \"aaa\".\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n^3) ๐Ÿช: O(1)\n ๐Ÿ‡ Solution: ๐Ÿ“ˆ Dynamic Programming โฐ: O(n^2) ๐Ÿช: O(n^2)\n\n var countSubstrings = function (s) {\n let count = 0;\n for (let i = 0; i \u003c s.length; i++) {\n count += exp\u0026(s, i, i);\n count += exp\u0026(s, i, i + 1);\n }\n return count;\n };\n function exp\u0026(s, left, right) {\n let count = 0;\n while (left \u003e= 0 \u0026\u0026 right \u003c s.length \u0026\u0026 s[left] === s[right]) {\n count++;\n left--;\n right++;\n }\n return count;\n }\n\n---\n\n## Tree\n## [Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)๐ŸŒฒ\n โ“: Given the root of a binary tree, return its maximum depth.\n A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.\n ๐Ÿฃ: 1๏ธโƒฃ Input: root = [3,9,20,null,null,15,7] Output: 3 2๏ธโƒฃ Input: root = [1,null,2] Output: 2 3๏ธโƒฃ Input: root = [] Output: 0 #4 Input: root = [0] Output: 1\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐ŸŒฒ DFS โฐ: O(n) ๐Ÿช: O(n)\n\n var maxDepth = function (root) {\n if (!root) return 0;\n return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;\n };\n\n## [Same Tree](https://leetcode.com/problems/same-tree/) #100๐ŸŒฒ\n โ“: Given the roots of two binary trees p \u0026 q, write a function to check if they are the same or not.\n Two binary trees are considered the same if they are structurally identical, \u0026 the nodes have the same value.\n ๐Ÿฃ: 1๏ธโƒฃ Input: p = [1,2,3], q = [1,2,3] Output: true 2๏ธโƒฃ Input: p = [1,2], q = [1,null,2] Output: false 3๏ธโƒฃ Input: p = [1,2,1], q = [1,1,2] Output: false\n\n ๐Ÿข Solution: ๐Ÿ”จ Brute Force โฐ: O(n) ๐Ÿช: O(n)\n ๐Ÿ‡ Solution: ๐ŸŒฒ DFS โฐ: O(n) ๐Ÿช: O(n)\n\n var isSameTree = function (p, q) {\n if (!p \u0026\u0026 !q) return true;\n if (!p || !q) return false;\n if (p.val !== q.val) return false;\n return isSameTree(p.left, q.left) \u0026\u0026 isSameTree(p.right, q.right);\n };\n\n## [Invert/Flip Binary Tree](https://leetcode.com/problems/invert-binary-tree/) #226๐ŸŒฒ\n โ“: Invert a binary tree.\n ๐Ÿฃ: 1๏ธโƒฃ Input: root = [4,2,7,1,3,6,9] Output: [4,7,2,9,6,3,1] 2๏ธโƒฃ Input: root = [2,1,3] Outpu","project_url":"https://awesome.ecosyste.ms/api/v1/projects/github.com%2Falishahusain%2Fnewt","html_url":"https://awesome.ecosyste.ms/projects/github.com%2Falishahusain%2Fnewt","lists_url":"https://awesome.ecosyste.ms/api/v1/projects/github.com%2Falishahusain%2Fnewt/lists"}