Time taken per hour for stoppage of Car

A car travels with an average speed of S km/h without any stoppage and with stoppage the speed of car reduces to an average of S1 km/h. The task is to find the time wasted per hour for the stoppage. 
Examples:

Input: S = 50, S1 = 30 
Output: 24 min
Input: S = 30, S1 = 10 
Output: 40 min 

Approach: Take the first example, 

Speed of car without any stoppage = 50 kmph i.e. 50 km in 60 min. 
Speed of car with stoppage = 30 kmph i.e. 30 km in 60 min.
Now, if there will be no stoppage then 30 km can be covered in 36 min. 
50 km –> 60 min 
30 km –> (60 / 50) * 30 = 36 min 
but it takes 60 min. 
So, time for stoppage per hour is 60 min – 36 min = 24 min. 

This can be calculated using the below formula: 

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the time taken
// per hour for stoppage
int numberOfMinutes(int S, int S1)
{

    int Min = 0;

    Min = ((S - S1) / floor(S)) * 60;

    return Min;
}

// Driver code
int main()
{
    int S = 30, S1 = 10;

    cout << numberOfMinutes(S, S1) << " min";

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Function to return the time taken
// per hour for stoppage
static int numberOfMinutes(int S, int S1)
{
    int Min = 0;

    Min = (int) (((S - S1) / Math.floor(S)) * 60);

    return Min;
}

// Driver code
public static void main(String[] args)
{
    int S = 30, S1 = 10;

    System.out.println(numberOfMinutes(S, S1) + " min");
}
}

// This code is contributed by Princi Singh

Python3

# Python3 implementation of the approach
import math

# Function to return the time taken
# per hour for stoppage
def numberOfMinutes(S, S1):

    Min = 0;

    Min = ((S - S1) / math.floor(S)) * 60;

    return int(Min);

# Driver code
if __name__ == '__main__':
    S, S1 = 30, 10;

    print(numberOfMinutes(S, S1), "min");

# This code is contributed by Rajput-Ji

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the time taken
// per hour for stoppage
static int numberOfMinutes(int S, int S1)
{
    int Min = 0;

    Min = (int) (((S - S1) /
                   Math.Floor((double)S)) * 60);

    return Min;
}

// Driver code
public static void Main()
{
    int S = 30, S1 = 10;

    Console.WriteLine(numberOfMinutes(S, S1) +
                                      " min");
}
}

// This code is contributed
// by Akanksha Rai

Javascript

<script>

// JavaScript implementation of the approach

// Function to return the time taken
// per hour for stoppage
function numberOfMinutes(S, S1)
{

    let Min = 0;

    Min = ((S - S1) / Math.floor(S)) * 60;

    return Min;
}

// Driver code

    let S = 30, S1 = 10;

    document.write(numberOfMinutes(S, S1) + " min");

// This code is contributed by Surbhi Tyagi.

</script>

Output:

40 min

Time Complexity: O(1), since there is only basic arithmetic that happens in constant time.

Auxiliary Space: O(1), since no extra space has been taken.