Program to find the head start in a race

Given the head start that A gives to B and C in a 100-meters race. The task is to find the head-start that B can give to C in the same race.

Examples:

Input: B = 10 meters, C = 28 meters
Output: 20 meters
B can give C a start of 20 meters.

Input: B = 20 meters, C = 50 meters
Output: 62 meters
B can give C a start of 62 meters.

Approach:

Total meters in a race = 100 meters. 
A is ahead of B by 10 meters. When A completed it’s 100 meters B completed it’s 90 meters. 
Similarly, A is ahead of C by 28 meters. When A completed it’s 100 meters C completed it’s 72 meters. 
Now, When B completed it’s 90 meters C completed it’s 72 meters. 
So when B completed it’s 100 meters C completed it’s 80 meters. 
–> (( C * 100) / B) 
–> (( 72 * 100) / 90) i.e 80 meters
So B can give C a start of 20 meters 

Below is the implementation of the above approach:

C++

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

// Function to find the B start to C
int Race(int B, int C)
{
    int result = 0;

    // When B completed it's 100 meter
    // then Completed meters by C is
    result = ((C * 100) / B);

    return 100 - result;
}

// Driver Code.
int main()
{
    int B = 10, C = 28;

    // When A completed it's 100 meter
    // Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;

    cout << Race(B, C) << " meters";

return 0;
}

Java

// Java implementation of above approach
public class GFG
{

// Function to find the B start to C
static int Race(int B, int C)
{
    int result = 0;

    // When B completed it's 100 meter
    // then Completed meters by C is
    result = ((C * 100) / B);

    return 100 - result;
}

// Driver Code
public static void main(String[] args)
{
    int B = 10;
    int C = 28;

    // When A completed it's 100 meter
    // Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;

    System.out.println(Race(B, C) + " meters");
}
}

// This code is contributed
// by ChitraNayal

Python3

# Python 3 implementation
# of above approach

# Function to find the
# B start to C
def Race(B, C):

    result = 0;

    # When B completed it's 100 meter
    # then Completed meters by C is
    result = ((C * 100) // B)

    return 100 - result

# Driver Code
if __name__ == "__main__":
    B = 10
    C = 28

    # When A completed it's 100 meter
    # Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;

    print(str(Race(B, C)) + " meters")

# This code is contributed
# by ChitraNayal

C#

// C# implementation of above approach
using System;
class GFG
{

// Function to find the B start to C
static int Race(int B, int C)
{
    int result = 0;

    // When B completed it's 100 meter
    // then Completed meters by C is
    result = ((C * 100) / B);

    return 100 - result;
}

// Driver Code
public static void Main()
{
    int B = 10;
    int C = 28;

    // When A completed it's 100 meter
    // Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;

    Console.Write(Race(B, C) + " meters");
}
}

// This code is contributed
// by ChitraNayal

PHP

<?php
// PHP implementation of above approach

// Function to find the B start to C
function Race($B, $C)
{
    $result = 0;

    // When B completed it's 100 meter
    // then Completed meters by C is
    $result = (($C * 100) / $B);

    return 100 - $result;
}

// Driver Code
$B = 10;
$C = 28;

// When A completed it's 100 meter
// Then completed meters of B and C is
$B = 100 - $B;
$C = 100 - $C;

echo Race($B, $C) . " meters";

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>

// Javascript implementation of above approach

// Function to find the B start to C
function Race(B, C)
{
    var result = 0;

    // When B completed it's 100 meter
    // then Completed meters by C is
    result = ((C * 100) / B);

    return 100 - result;
}

// Driver Code
var B = 10, C = 28;

// When A completed it's 100 meter
// Then completed meters of B and C is
B = 100 - B;
C = 100 - C;

document.write(Race(B, C) + " meters");

// This code is contributed by itsok

</script>

Output:

20 meters

Time Complexity: O(1)

Auxiliary Space: O(1)