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

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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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

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C#

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// 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.

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PHP

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<?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.
?>

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Output:

8 Hours


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