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

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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:

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

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

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