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++ 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;
} |
// 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. |
# 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 |
// 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 |
<?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 contributed // by Shivi_Aggarwal ?> |
<script> // javascript implementation // of above approach // Function to calculate the time
function calTime(arr , n)
{
var work = 0;
for (i = 0; i < n; i++)
work += 1 / arr[i];
return 1 / work;
}
// Driver Code
var arr = [ 6.0, 3.0, 4.0 ];
var n = arr.length;
document.write(calTime(arr, n).toFixed(5) + " Hours" );
// This code is contributed by Rajput-Ji. </script> |
1.33333 Hours
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.