739. Daily Temperatures

1. 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.

2. Example

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]

3. Constraints

• 1 <= temperatures.length <= $10^5$
• 30 <= temperatures[i] <= 100

4. Solutions

Monotonic Stack

n = temperatures.size()
Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
    vector<int> dailyTemperatures(const vector<int> &temperatures) {
        vector<int> wait_days(temperatures.size(), 0);
        stack<int, vector<int>> indexes;
        indexes.push(0);

        for (int i = 1; i < temperatures.size(); ++i) {
            while (!indexes.empty() && temperatures[i] > temperatures[indexes.top()]) {
                int index = indexes.top();
                indexes.pop();
                wait_days[index] = i - index;
            }

            indexes.push(i);
        }

        return wait_days;
    }
};