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++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the timefloat calTime(float arr[], int n){ float work = 0; for (int i = 0; i < n; i++) work += 1 / arr[i]; return 1 / work;}// Driver Codeint main(){ float arr[] = { 6.0, 3.0, 4.0 }; int n = sizeof(arr) / sizeof(arr[0]); cout << calTime(arr, n) << " Hours"; return 0;} |
Java
// Java implementation // of above approachimport java.io.*;class GFG { // Function to calculate the timestatic double calTime(double arr[], int n){ double work = 0; for (int i = 0; i < n; i++) work += 1 / arr[i]; return 1 / work;}// Driver Codepublic 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
# 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#
// C# implementation // of above approachusing System;class GFG { // Function to calculate the timestatic 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 Codepublic 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// 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?> |
Javascript
<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> |
Output:
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.
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