# 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 beb / 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 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; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// 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. ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## C#

`// 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 ` |

*chevron_right*

*filter_none*

## 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 contribuuted ` `// by Shivi_Aggarwal ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

1.33333 Hours

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

## Recommended Posts:

- Minimum operations of the given type required to make a complete graph
- Minimum time required to fill a cistern using N pipes
- Time taken by two persons to meet on a circular track
- Program to find the time remaining for the day to complete
- Number of ways to arrange 2*N persons on the two sides of a table with X and Y persons on opposite sides
- Changing One Clock Time to Other Time in Minimum Number of Operations
- Minimum time to reach a point with +t and -t moves at time t
- Time required to meet in equilateral triangle
- Minimum Players required to win the game
- Minimum number of given operation required to convert n to m
- Minimum operations required to change the array such that |arr[i] - M| <= 1
- Minimum number of changes required to make the given array an AP
- Minimum number of operations required to reduce N to 1
- Minimum number operations required to convert n to m | Set-2
- Minimum inversions required so that no two adjacent elements are same

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.