---
title: '739. Daily Temperatures'
description: 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 i^th day to get a warmer temperature. If there is no future day for which this is possible, keep answer[i] == 0 instead
sidebar:
label: 'Daily Temperatures'
badge: 'Medium'
---
Array
### 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 <= 10^5`
- `30 <= temperatures[i] <= 100`
## Approach
```mermaid
flowchart TD
S(["dailyTemperatures(temperatures)"]) --> I["ans = zeros(n), stack = [] — indices of days still waiting"]
I --> L{"more (i, temp)?"}
L -- no --> E(["return ans"])
L -- yes --> W{"stack non-empty and temperatures[stack[-1]] < temp?"}
W -- yes --> P["prev = stack.pop(); ans[prev] = i - prev — today resolves that day"]
P --> W
W -- no --> A["stack.append(i)"]
A --> L
```
## Solution
```py
class Solution:
def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
ans = [0] * len(temperatures)
stack = []
for i, temp in enumerate(temperatures):
while stack and temperatures[stack[-1]] < temp:
prev = stack.pop()
ans[prev] = i - prev
stack.append(i)
return ans
```
## Explanation