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Count of days remaining for the next day with higher temperature

  • Difficulty Level : Easy
  • Last Updated : 10 Jun, 2021

Given a list arr[] of everyday temperatures. For each day, the task is to find the count of days remaining for the next day with warmer temperatures. If there is no such day for which warmer temperature is possible then print -1.
Examples:

Input: arr[] = {73, 74, 75, 71, 69, 72, 76, 73} 
Output: {1, 1, 4, 2, 1, 1, -1, -1} 
Explanation: 
For 73 temperature, next warmer temperature is 74 which at a distance 1 
For 74 temperature, next warmer temperature is 75 which at a distance 1 
For 75 temperature, next warmer temperature is 76 which at a distance 4 
For 71 temperature, next warmer temperature is 72 which at a distance 2 
For 69 temperature, next warmer temperature is 72 which at a distance 1 
For 72 temperature, next warmer temperature is 76 which at a distance 1 
For 76 temperature, there is no valid next warmer day 
For 73 temperature, there is no valid next warmer day
Input: arr[] = {75, 74, 73, 72} 
Output: {-1, -1, -1, -1}

Naive Approach: The idea is to iterate for each possible pairs of the array and check the next greater temperatures for each current element.
Time Complexity: O(N2
Auxiliary Space: O(1)
Efficient Approach: This problem basically asks to find how far is the current index from the index of next greater temperature to the temperature at the current index. The most optimal way to solve this problem is by making use of a stack. Below are the steps:

  1. Iterate over the everyday temperature of the given array arr[] using the current index.
  2. If the stack is empty, push the current index to the stack.
  3. If the stack is not empty then do the following: 
    • If the temperature at the current index is lesser than the temperature of the index at top of the stack, push the current index.
    • If the temperature at the current index is greater than the temperature of the index at top of the stack, then update the no of days to wait for warmer temperature as:

 
 

current index – index at top of the stack



 

  • Pop the stack once the number of days has been updated in the output list.
  1. Repeat the above steps for all the indices in the stack that are lesser than the temperature at the current index.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to determine how many days
// required to wait for the next
// warmer temperature
void dailyTemperatures(vector<int>& T)
{
    int n = T.size();
 
    // To store the answer
    vector<int> daysOfWait(n, -1);
    stack<int> s;
 
    // Traverse all the temperatures
    for (int i = 0; i < n; i++) {
 
        // Check if current index is the
        // next warmer temperature of
        // any previous indexes
        while (!s.empty()
               && T[s.top()] < T[i]) {
 
            daysOfWait[s.top()]
                = i - s.top();
 
            // Pop the element
            s.pop();
        }
 
        // Push the current index
        s.push(i);
    }
 
    // Print waiting days
    for (int i = 0; i < n; i++) {
        cout << daysOfWait[i] << " ";
    }
}
 
// Driver Code
int main()
{
    // Given temperatures
    vector<int> arr{ 73, 74, 75, 71,
                     69, 72, 76, 73 };
 
    // Function Call
    dailyTemperatures(arr);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
import java.lang.*;
 
class GFG{
 
// Function to determine how many days
// required to wait for the next
// warmer temperature
static void dailyTemperatures(int[] T)
{
    int n = T.length;
 
    // To store the answer
    int[] daysOfWait = new int[n];
    Arrays.fill(daysOfWait, -1);
     
    Stack<Integer> s = new Stack<>();
 
    // Traverse all the temperatures
    for(int i = 0; i < n; i++)
    {
         
        // Check if current index is the
        // next warmer temperature of
        // any previous indexes
        while (!s.isEmpty() && T[s.peek()] < T[i])
        {
            daysOfWait[s.peek()] = i - s.peek();
             
            // Pop the element
            s.pop();
        }
         
        // Push the current index
        s.push(i);
    }
 
    // Print waiting days
    for(int i = 0; i < n; i++)
    {
        System.out.print(daysOfWait[i] + " ");
    }
}
 
// Driver Code
public static void main (String[] args)
{
     
    // Given temperatures
    int[] arr = { 73, 74, 75, 71,
                  69, 72, 76, 73 };
     
    // Function call
    dailyTemperatures(arr);
}
}
 
// This code is contributed by offbeat

Python3




# Python3 program for the above approach
 
# Function to determine how many days
# required to wait for the next
# warmer temperature
def dailyTemperatures(T):
 
    n = len(T)
 
    # To store the answer
    daysOfWait = [-1] * n
    s = []
 
    # Traverse all the temperatures
    for i in range(n):
 
        # Check if current index is the
        # next warmer temperature of
        # any previous indexes
        while(len(s) != 0 and
              T[s[-1]] < T[i]):
            daysOfWait[s[-1]] = i - s[-1]
 
            # Pop the element
            s.pop(-1)
 
        # Push the current index
        s.append(i)
 
    # Print waiting days
    for i in range(n):
        print(daysOfWait[i], end = " ")
 
# Driver Code
 
# Given temperatures
arr = [ 73, 74, 75, 71,
        69, 72, 76, 73 ]
 
# Function call
dailyTemperatures(arr)
 
# This code is contributed by Shivam Singh

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to determine how many days
// required to wait for the next
// warmer temperature
static void dailyTemperatures(int[] T)
{
    int n = T.Length;
 
    // To store the answer
    int[] daysOfWait = new int[n];
    for(int i = 0; i < n; i++)
        daysOfWait[i] = -1;
     
    Stack<int> s = new Stack<int>();
 
    // Traverse all the temperatures
    for(int i = 0; i < n; i++)
    {
         
        // Check if current index is the
        // next warmer temperature of
        // any previous indexes
        while (s.Count != 0 && T[s.Peek()] < T[i])
        {
            daysOfWait[s.Peek()] = i - s.Peek();
             
            // Pop the element
            s.Pop();
        }
         
        // Push the current index
        s.Push(i);
    }
 
    // Print waiting days
    for(int i = 0; i < n; i++)
    {
        Console.Write(daysOfWait[i] + " ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given temperatures
    int[] arr = { 73, 74, 75, 71,
                  69, 72, 76, 73 };
     
    // Function call
    dailyTemperatures(arr);
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
 
 
// JavaScript program for the above approach
 
// Function to determine how many days
// required to wait for the next
// warmer temperature
function dailyTemperatures(T)
{
    var n = T.length;
 
    // To store the answer
    var daysOfWait = Array(n).fill(-1);
    var s = [];
 
    // Traverse all the temperatures
    for (var i = 0; i < n; i++) {
 
        // Check if current index is the
        // next warmer temperature of
        // any previous indexes
        while (s.length!=0
               && T[s[s.length-1]] < T[i]) {
 
            daysOfWait[s[s.length-1]]
                = i - s[s.length-1];
 
            // Pop the element
            s.pop();
        }
 
        // Push the current index
        s.push(i);
    }
 
    // Print waiting days
    for (var i = 0; i < n; i++) {
        document.write( daysOfWait[i] + " ");
    }
}
 
// Driver Code
// Given temperatures
var arr = [73, 74, 75, 71,
                 69, 72, 76, 73 ];
// Function Call
dailyTemperatures(arr);
 
 
</script>
Output: 
1 1 4 2 1 1 -1 -1

Time Complexity: O(N) 
Auxiliary Space: O(N)
 

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