Total number of days taken to complete the task if after certain days one person leaves

Given that person A takes a days to do a certain piece of work while person B takes b days to do the same work. If A and B started the work together and n days before the completion of work, A leaves the work. Find the total number of days taken to complete work.

Examples:

Input: a = 10, b = 20, n = 5 
Output: 10

Input: a = 5, b = 15, n = 3 
Output: 6  

Approach: Let D be the total number of days to complete the work. A and B work together for D – n days. So,  

(D – n) * (1/a + 1/b) + (n) * (1/b) = 1 
D * (1/a + 1/b) – (n/a) – (n/b) + (n/b) = 1 
D * (1/a + 1/b) = (n + a) / a 
D = b * (n + a) / (a + b)

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
// number of days required
int numberOfDays(int a, int b, int n)
{
    int Days = b * (n + a) / (a + b);

    return Days;
}

// Driver code
int main()
{
    int a = 10, b = 20, n = 5;

    cout << numberOfDays(a, b, n);

    return 0;
}

Java

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

class GFG
{

// Function to return the
// number of days required
static int numberOfDays(int a, int b, int n)
{
    int Days = b * (n + a) / (a + b);

    return Days;
}

// Driver code
public static void main (String[] args)
{

    int a = 10, b = 20, n = 5;
    System.out.println (numberOfDays(a, b, n));

}
}

// This code is contributed by jit_t

Python3

# Python3 implementation of the approach

# Function to return the
# number of days required
def numberOfDays(a, b, n):

    Days = b * (n + a) // (a + b)

    return Days

# Driver code
if __name__ == "__main__" :

    a = 10
    b = 20
    n = 5

    print(numberOfDays(a, b, n))

# This code is contributed by rrrtnx

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the
    // number of days required
    static int numberOfDays(int a, int b, int n)
    {
        int Days = b * (n + a) / (a + b);

        return Days;
    }

    // Driver code
    public static void Main ()
    {

        int a = 10, b = 20, n = 5;
        Console.WriteLine(numberOfDays(a, b, n));

    }
}

// This code is contributed by AnkitRai01

Javascript

<script>
// javascript implementation of the approach

    // Function to return the
    // number of days required
    function numberOfDays(a , b , n)
    {
        var Days = b * (n + a) / (a + b);
        return Days;
    }

    // Driver code
    var a = 10, b = 20, n = 5;
    document.write(numberOfDays(a, b, n));

// This code is contributed by Rajput-Ji
</script>

Output:

10

Time Complexity: O(1), since there is only basic arithmetic operation.

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