# Minimum time required to fill a cistern using N pipes

Last Updated : 30 Mar, 2023

Given the time required by a total of N+1 pipes where N pipes are used to fill the Cistern and a single pipe is used to empty the Cistern. The task is to Calculate the amount of time in which the Cistern will get filled if all the N+1 pipes are opened together.

Examples

```Input: n = 2,
pipe1 = 12 hours, pipe2 = 14 hours,
emptypipe = 30 hours
Output: 8 hours

Input: n = 1,
pipe1 = 12 hours
emptypipe = 18 hours
Output: 36 hours ```

Approach:

• If a pipe1 can fill a cistern in â€˜nâ€™ hours, then in 1 hour, the pipe1 will able to fill â€˜1/nâ€™ Cistern.
• Similarly If a pipe2 can fill a cistern in â€˜mâ€™ hours, then in one hour, the pipe2 will able to fill â€˜1/mâ€™ Cistern.
• So onâ€¦. for other pipes.

So, total work done in filling a Cistern by N pipes in 1 hours is

1/n + 1/m + 1/pâ€¦â€¦ + 1/z
Where n, m, p â€¦.., z are the number of hours taken by each pipes respectively.

The result of the above expression will be the part of work done by all pipes together in 1 hours, letâ€™s say a / b.
To calculate the time taken to fill the cistern will be b / a.

Consider an example of two pipes:

Time taken by 1st pipe to fill the cistern = 12 hours
Time taken by 2nd pipe to fill the cistern = 14 hours
Time taken by 3rd pipe to empty the cistern = 30 hours
Work done by 1st pipe in 1 hour = 1/12
Work done by 2nd pipe in 1 hour = 1/14
Work done by 3rd pipe in 1 hour = – (1/30) as it empty the pipe.
So, total work done by all the pipes in 1 hour is
=> ( 1 / 12 + 1/ 14 ) – (1 / 30)
=> ((7 + 6 ) / (84)) – (1 / 30)
=> ((13) / (84)) – (1 / 30)
=> 51 / 420
So, to Fill the cistern time required will be 420 / 51 i.e 8 hours Approx.

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach` `#include ` `using` `namespace` `std;`   `// Function to calculate the time` `float` `Time(``float` `arr[], ``int` `n, ``int` `Emptypipe)` `{`   `    ``float` `fill = 0;` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``fill += 1 / arr[i];`   `    ``fill = fill - (1 / (``float``)Emptypipe);`   `    ``return` `1 / fill;` `}`   `// Driver Code` `int` `main()` `{` `    ``float` `arr[] = { 12, 14 };` `    ``float` `Emptypipe = 30;` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``cout << ``floor``(Time(arr, n, Emptypipe)) << ``" Hours"``;`   `    ``return` `0;` `}`

## Java

 `// Java implementation of` `// above approach` `import` `java.io.*;`   `class` `GFG` `{` `    `  `// Function to calculate the time` `static` `float` `Time(``float` `arr[], ``int` `n,` `                  ``float` `Emptypipe)` `{` `    ``float` `fill = ``0``;` `    ``for` `(``int` `i = ``0``; i < n; i++)` `        ``fill += ``1` `/ arr[i];`   `    ``fill = fill - (``1` `/ (``float``)Emptypipe);`   `    ``return` `1` `/ fill;` `}`   `// Driver Code` `public` `static` `void` `main (String[] args) ` `{` `    ``float` `arr[] = { ``12``, ``14` `};` `    ``float` `Emptypipe = ``30``;` `    ``int` `n = arr.length;` `    `  `    ``System.out.println((``int``)(Time(arr, n, ` `                        ``Emptypipe)) + ``" Hours"``);` `}` `}`   `// This code is contributed` `// by inder_verma.`

## Python3

 `# Python3 implementation of ` `# above approach `   `# Function to calculate the time ` `def` `Time(arr, n, Emptypipe) :`   `    ``fill ``=` `0` `    ``for` `i ``in` `range``(``0``,n) :` `        ``fill ``+``=` `(``1` `/` `arr[i]) `   `    ``fill ``=` `fill ``-` `(``1` `/` `float``(Emptypipe)) `   `    ``return` `int``(``1` `/` `fill) `     `# Driver Code ` `if` `__name__``=``=``'__main__'``:` `    ``arr ``=` `[ ``12``, ``14` `] ` `    ``Emptypipe ``=` `30` `    ``n ``=` `len``(arr) ` `    ``print``((Time(arr, n, Emptypipe)) ` `          ``, ``"Hours"``)`   `# This code is contributed by` `# Smitha Dinesh Semwal`

## C#

 `// C# implementation of` `// above approach` `using` `System;`   `class` `GFG` `{` `    `  `// Function to calculate the time` `static` `float` `Time(``float` `[]arr, ``int` `n,` `                  ``float` `Emptypipe)` `{` `    ``float` `fill = 0;` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``fill += 1 / arr[i];`   `    ``fill = fill - (1 / (``float``)Emptypipe);`   `    ``return` `1 / fill;` `}`   `// Driver Code` `public` `static` `void` `Main () ` `{` `    ``float` `[]arr = { 12, 14 };` `    ``float` `Emptypipe = 30;` `    ``int` `n = arr.Length;` `    `  `    ``Console.WriteLine((``int``)(Time(arr, n, ` `                             ``Emptypipe)) + ` `                                ``" Hours"``);` `}` `}`   `// This code is contributed` `// by inder_verma.`

## PHP

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## Javascript

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Output:

`8 Hours`

Time Complexity: O(1)

Auxiliary Space: O(1)