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 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++
#include <bits/stdc++.h>
using namespace std;
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;
}
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
import java.io.*;
class GFG
{
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;
}
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" );
}
}
|
Python3
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)
if __name__ = = '__main__' :
arr = [ 12 , 14 ]
Emptypipe = 30
n = len (arr)
print ((Time(arr, n, Emptypipe))
, "Hours" )
|
C#
using System;
class GFG
{
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;
}
public static void Main ()
{
float []arr = { 12, 14 };
float Emptypipe = 30;
int n = arr.Length;
Console.WriteLine(( int )(Time(arr, n,
Emptypipe)) +
" Hours" );
}
}
|
PHP
<?php
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 ;
}
$arr = array ( 12, 14 );
$Emptypipe = 30;
$n = count ( $arr );
echo (int)T_ime( $arr , $n ,
$Emptypipe ) . " Hours" ;
?>
|
Javascript
<script>
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;
}
var arr = [ 12, 14 ];
var Emptypipe = 30;
var n = arr.length;
document.write(Math.floor(
Time(arr, n, Emptypipe)) + " Hours" );
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Last Updated :
30 Mar, 2023
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