https://github.com/emahtab/daily-temperatures
Find the days after which there will be a warmer day
https://github.com/emahtab/daily-temperatures
leetcode problem-solving stack
Last synced: 20 days ago
JSON representation
Find the days after which there will be a warmer day
- Host: GitHub
- URL: https://github.com/emahtab/daily-temperatures
- Owner: eMahtab
- Created: 2020-06-13T18:54:38.000Z (about 5 years ago)
- Default Branch: master
- Last Pushed: 2024-10-12T05:26:43.000Z (9 months ago)
- Last Synced: 2025-03-28T01:42:08.670Z (4 months ago)
- Topics: leetcode, problem-solving, stack
- Homepage:
- Size: 6.84 KB
- Stars: 1
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
Awesome Lists containing this project
README
# Daily Temperatures
## https://leetcode.com/problems/daily-temperatures
Given a list of daily temperatures T, return a list such that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put 0 instead.
```
For example, given the list of temperatures T = [73, 74, 75, 71, 69, 72, 76, 73],
your output should be [1, 1, 4, 2, 1, 1, 0, 0].
```
**Note: The length of temperatures will be in the range [1, 30000]. Each temperature will be an integer in the range [30, 100].**
### Implementation : Naive
```java
class Solution {
public int[] dailyTemperatures(int[] T) {
if(T == null || T.length == 0)
return T;
for(int i = 0; i < T.length; i++) {
int daysToWait = 1;
boolean warmerDay = false;
for(int j = i + 1; j < T.length; j++) {
if(T[j] > T[i]) {
warmerDay = true;
T[i] = daysToWait;
break;
}
daysToWait++;
}
if(!warmerDay)
T[i] = 0;
}
return T;
}
}
```
## Implementation 2 : Using stack (Iterating from left to right)
```java
class Solution {
public int[] dailyTemperatures(int[] temperatures) {
int n = temperatures.length;
if(temperatures == null || n == 0)
return new int[0];
if(n == 1)
return new int[]{0};
Stack stack = new Stack<>();
int[] result = new int[n];
for(int i = 0; i < n; i++) {
while(!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]) {
int index = stack.pop();
result[index] = i - index;
}
stack.push(i);
}
while(!stack.isEmpty()) {
int index = stack.pop();
result[index] = 0;
}
return result;
}
}
```
## Implementation 3 : Using stack (Iterating from right to left)
```java
class Solution {
public int[] dailyTemperatures(int[] temperatures) {
int n = temperatures.length;
if(temperatures == null || n == 0)
return new int[0];
if(n == 1)
return new int[]{0};
Stack stack = new Stack<>();
int[] result = new int[n];
result[n-1] = 0;
stack.push(n-1);
for(int i = n-2; i >= 0; i--) {
while(!stack.isEmpty() && temperatures[stack.peek()] <= temperatures[i]) {
stack.pop();
}
if(stack.isEmpty())
result[i] = 0;
else
result[i] = stack.peek() - i;
stack.push(i);
}
return result;
}
}
```