# Minimum time required to fill a cistern using N pipes

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• If a pipe1 can fill a cistern in ‘n’ hours, then in 1 hour, the pipe1 will able to fill ‘1/n’ Cistern.
• Similarly If a pipe2 can fill a cistern in ‘m’ hours, then in one hour, the pipe2 will able to fill ‘1/m’ Cistern.
• So on…. for other pipes.

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

1/n + 1/m + 1/p…… + 1/z
Where n, m, p ….., 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, let’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 3nd 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

 ` `

Output:

```8 Hours
```

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