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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