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) {
        const int n = temperatures.size();
        vector<int> wait_days(n, 0), to_solve;
        for (int i = 0; i < n; ++i) {
            while (!to_solve.empty() && temperatures[i] > temperatures[to_solve.back()]) {
                wait_days[to_solve.back()] = i - to_solve.back();
                to_solve.pop_back();
            }

            to_solve.push_back(i);
        }

        return wait_days;
    }
};
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