https://github.com/emahtab/container-with-most-water
Container with most water problem
https://github.com/emahtab/container-with-most-water
container-with-most-water leetcode leetcode-java problem-solving two-pointers
Last synced: 3 months ago
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Container with most water problem
- Host: GitHub
- URL: https://github.com/emahtab/container-with-most-water
- Owner: eMahtab
- Created: 2019-12-25T05:36:11.000Z (over 5 years ago)
- Default Branch: master
- Last Pushed: 2020-04-01T07:14:27.000Z (over 5 years ago)
- Last Synced: 2025-02-02T03:26:11.345Z (5 months ago)
- Topics: container-with-most-water, leetcode, leetcode-java, problem-solving, two-pointers
- Homepage:
- Size: 21.5 KB
- Stars: 1
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# Container with most water
## https://leetcode.com/problems/container-with-most-water
Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container and n is at least 2.
The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
**This a beautiful question showing, how to use the two pointer approach for better runtime**
### Implementation 1 : O(n^2)
```java
class Solution {
public int maxArea(int[] height) {
if(height == null || height.length == 0)
return 0;
int mostWater = Integer.MIN_VALUE;
for(int i = 0; i < height.length - 1; i++) {
for(int j = i + 1; j < height.length; j++) {
mostWater = Math.max(mostWater, Math.min(height[i], height[j]) * (j-i));
}
}
return mostWater;
}
}
```
Above implementation have runtime complexity of O(n^2) and space complexity of O(1)
```
Runtime Complexity = O(n^2)
Space Complexity = O(1)
```
### Implementation 2 : O(n)
```java
class Solution {
public int maxArea(int[] height) {
if(height == null || height.length == 0)
return 0;
int left = 0;
int right = height.length - 1;
int mostWater = Integer.MIN_VALUE;
while(left < right) {
int distance = right - left;
int waterStored = Math.min(height[left], height[right]) * distance;
mostWater = Math.max(mostWater, waterStored);
if(height[left] <= height[right]) {
left++;
} else {
right--;
}
}
return mostWater;
}
}
```
Above implementation have runtime complexity of O(n) and space complexity of O(1)
```
Runtime Complexity = O(n)
Space Complexity = O(1)
```