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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]


  • 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;
    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

    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;
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