Minimum time required to complete a work by N persons together
Last Updated :
16 Jun, 2022
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++
#include <bits/stdc++.h>
using namespace std;
float calTime( float arr[], int n)
{
float work = 0;
for ( int i = 0; i < n; i++)
work += 1 / arr[i];
return 1 / work;
}
int main()
{
float arr[] = { 6.0, 3.0, 4.0 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << calTime(arr, n) << " Hours" ;
return 0;
}
|
Java
import java.io.*;
class GFG
{
static double calTime( double arr[], int n)
{
double work = 0 ;
for ( int i = 0 ; i < n; i++)
work += 1 / arr[i];
return 1 / work;
}
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" );
}
}
|
Python3
def calTime(arr, n):
work = 0
for i in range (n):
work + = 1 / arr[i]
return 1 / work
arr = [ 6.0 , 3.0 , 4.0 ]
n = len (arr)
print (calTime(arr, n), "Hours" )
|
C#
using System;
class GFG
{
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);
}
public static void Main ()
{
double []arr = { 6.0, 3.0, 4.0 };
int n = arr.Length;
Console.Write(calTime(arr, n) +
" Hours" );
}
}
|
PHP
<?php
function calTime(& $arr , $n )
{
$work = 0;
for ( $i = 0; $i < $n ; $i ++)
$work += 1 / $arr [ $i ];
return 1 / $work ;
}
$arr = array (6.0, 3.0, 4.0);
$n = sizeof( $arr );
echo calTime( $arr , $n );
echo " Hours" ;
?>
|
Javascript
<script>
function calTime(arr , n)
{
var work = 0;
for (i = 0; i < n; i++)
work += 1 / arr[i];
return 1 / work;
}
var arr = [ 6.0, 3.0, 4.0 ];
var n = arr.length;
document.write(calTime(arr, n).toFixed(5) + " Hours" );
</script>
|
Time Complexity: O(n), to iterate over the array
Auxiliary Space: O(1)
Note: Here the input array contains hours, it can be days, minutes……so on.
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