Minimum time required to fill a cistern using N pipes

Last Updated : 13 Apr, 2021

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 ‘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 <bits/stdc++.h>
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

<?php
// PHP implementation of above approach
 
// Function to calculate the time
function T_ime($arr, $n, $Emptypipe)
{
    $fill = 0;
    for ($i = 0; $i < $n; $i++)
        $fill += 1 / $arr[$i];
 
    $fill = $fill - (1 / $Emptypipe);
 
    return 1 / $fill;
}
 
// Driver Code
$arr = array( 12, 14 );
$Emptypipe = 30;
$n = count($arr);
 
echo (int)T_ime($arr, $n,
                $Emptypipe) . " Hours";
 
// This code is contributed
// by inder_verma.
?>

Javascript

<script>
 
// Javascript implementation of above approach
 
// Function to calculate the time
function Time(arr, n, Emptypipe)
{
    var fill = 0;
    for(var i = 0; i < n; i++)
        fill += 1 / arr[i];
 
    fill = fill - (1 / Emptypipe);
 
    return 1 / fill;
}
 
// Driver Code
var arr = [ 12, 14 ];
var Emptypipe = 30;
var n = arr.length;
 
document.write(Math.floor(
    Time(arr, n, Emptypipe)) + " Hours");
     
// This code is contributed by itsok
 
</script>

Output:

8 Hours

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Minimum time required to fill a cistern using N pipes

C++

Java

Python3

C#

PHP

Javascript

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