Problem Description

You are given an array of integers temperatures where temperatures[i] represents the daily temperatures on the ith day.

Return an array result where result[i] is the number of days after the ith day before a warmer temperature appears on a future day. If there is no day in the future where a warmer temperature will appear for the ith day, set result[i] to 0 instead.

Example 1:

Input: temperatures = [30,38,30,36,35,40,28]
Output: [1,4,1,2,1,0,0]

Example 2:

Input: temperatures = [22,21,20]
Output: [0,0,0]

Constraints:

• 1 <= temperatures.length <= 1000
• 1 <= temperatures[i] <= 100

Approach

The provided solution uses nested loops (a brute-force like approach) combined with a stack to find the next warmer day. For each day, it scans forward until it finds a strictly greater temperature, adding intermediate temperatures to a stack along the way. The size of the stack determines the number of days waited.

Solution

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        Stack<Integer> st = new Stack<>();
        int[] array = new int[temperatures.length];  
        int n = temperatures.length;
/*
        for(int i=0;i<n;i++){
            int index = i+1;
            while(index<n){
                st.push(index);
                if(st.peek()<=temperatures[i]){
                    continue;
                }   
            }
        }
*/
        for(int i=0;i<n-1;i++){
            int index = i+1;
            while(index<n && temperatures[i]>=temperatures[index]){
                st.push(temperatures[index]);
                index++;
                
            }
            if(index<n && temperatures[i]<temperatures[index]){
                array[i] = st.size()+1;
            }
            else{
                array[i] = 0;

            }
            st = new Stack<>();

        }
        return array;
    }
}