Minimum time required to complete a work by N persons together

Given the no. of working hours of N people individually to complete a certain piece of work. The task is to find the number of hours they will take when all work together.

Examples:

Input: n = 2, a = 6.0, b = 3.0
Output: 2 Hours

Input: n = 3, a = 6.0, b = 3.0, c = 4.0
Output: 1.33333 Hours



Solution:

  • If a person can do a piece of work in ‘n’ days, then in one day, the person will do ‘1/n’ work.
  • Similarly If a person can do a piece of work in ‘m’ days, then in one day, the person will do ‘1/m’ work.
  • So on…. for other persons.

So, total work done by N persons in 1 day is

1/n + 1/m + 1/p…… + 1/z
Where n, m, p ….., z are the number of days taken by each person respectively.

The result of the above expression will be the part of work done by all person together in 1 day, let’s say a / b.
To calculate the time taken to complete the whole work will be b / a.

Consider an example of two persons:

Time taken by 1st person to complete a work = 6 hours
Time taken by 2nd person to complete the same work = 2 hours

Work done by 1st person in 1 hour = 1/6
Work done by 2nd person in 1 hour = 1/2
So, total work done by them in 1 hour is
=> 1 / 6 + 1/ 2 
=> (2 + 6) / (2 * 6)
=> 8 / 12

So, to complete the whole work, the time taken will be 12/8.

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate the time
float calTime(float arr[], int n)
{
  
    float work = 0;
    for (int i = 0; i < n; i++)
        work += 1 / arr[i];
  
    return 1 / work;
}
  
// Driver Code
int main()
{
    float arr[] = { 6.0, 3.0, 4.0 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << calTime(arr, n) << " Hours";
  
    return 0;
}

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Java

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// Java implementation 
// of above approach
import java.io.*;
  
class GFG 
{
      
// Function to calculate the time
static double calTime(double arr[], int n)
{
    double work = 0;
    for (int i = 0; i < n; i++)
        work += 1 / arr[i];
  
    return 1 / work;
}
  
// Driver Code
public static void main (String[] args) 
{
    double arr[] = { 6.0, 3.0, 4.0 };
    int n = arr.length;
      
    System.out.println(calTime(arr, n) +
                              " Hours");
}
}
  
// This code is contributed
// by inder_verma.

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Python3

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# Python3 implementation of 
# above approach 
  
# Function to calculate the time 
def calTime(arr, n): 
  
    work = 0
    for i in range(n):
        work += 1 / arr[i]
  
    return 1 / work
      
# Driver Code 
arr = [ 6.0, 3.0, 4.0
n = len(arr)
  
print(calTime(arr, n), "Hours")
  
# This code is contributed
# by Sanjit_Prasad

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C#

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// C# implementation 
// of above approach
using System;
class GFG 
{
      
// Function to calculate the time
static double calTime(double []arr, 
                      int n)
{
    double work = 0;
    for (int i = 0; i < n; i++)
        work += 1 / arr[i];
  
    return Math.Round(1 / work, 5);
}
  
// Driver Code
public static void Main () 
{
    double []arr = { 6.0, 3.0, 4.0 };
    int n = arr.Length;
      
    Console.Write(calTime(arr, n) +
                         " Hours");
}
}
  
// This code is contributed by Smitha

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PHP

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<?php
// PHP implementation of above approach
  
// Function to calculate the time 
function calTime(&$arr, $n
    $work = 0; 
    for ($i = 0; $i < $n; $i++) 
        $work += 1 / $arr[$i]; 
  
    return 1 / $work
  
// Driver Code 
$arr = array(6.0, 3.0, 4.0); 
$n = sizeof($arr); 
  
echo calTime($arr, $n);
echo " Hours";
  
// This code is contribuuted
// by Shivi_Aggarwal
?>

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Output:

1.33333 Hours


Note:
Here the input array contains hours, it can be days, minutes……so on.



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