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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