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# Daily Temperatures LeetCode Solution – Queslers

## Problem – Daily Temperatures

Given an array of integers `temperatures` represents the daily temperatures, return an array `answer` such that `answer[i]` is the number of days you have to wait after the `ith` day to get a warmer temperature. If there is no future day for which this is possible, keep `answer[i] == 0` instead.

Example 1:

``````Input: temperatures = [73,74,75,71,69,72,76,73]
Output: [1,1,4,2,1,1,0,0]``````

Example 2:

``````Input: temperatures = [30,40,50,60]
Output: [1,1,1,0]``````

Example 3:

``````Input: temperatures = [30,60,90]
Output: [1,1,0]``````

Constraints:

• `1 <= temperatures.length <= 105`
• `30 <= temperatures[i] <= 100`

### Daily Temperatures LeetCode Solution in Java

``````public int[] dailyTemperatures(int[] temperatures) {
Stack<Integer> stack = new Stack<>();
int[] ret = new int[temperatures.length];
for(int i = 0; i < temperatures.length; i++) {
while(!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]) {
int idx = stack.pop();
ret[idx] = i - idx;
}
stack.push(i);
}
return ret;
}
``````

### Daily Temperatures LeetCode Solution in Python

``````  def dailyTemperatures(self, T):
ans = [0] * len(T)
stack = []
for i, t in enumerate(T):
while stack and T[stack[-1]] < t:
cur = stack.pop()
ans[cur] = i - cur
stack.append(i)

return ans
``````

### Daily Temperatures LeetCode Solution in C++

``````vector<int> dailyTemperatures(vector<int>& temperatures) {
vector<int> res(temperatures.size());
for (int i = temperatures.size() - 1; i >= 0; --i) {
int j = i+1;
while (j < temperatures.size() && temperatures[j] <= temperatures[i]) {
if (res[j] > 0) j = res[j] + j;
else j = temperatures.size();
}
// either j == size || temperatures[j] > temperatures[i]
if (j < temperatures.size()) res[i] = j - i;
}
return res;
}``````
##### Daily Temperatures LeetCode Solution Review:

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