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++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the timefloat 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 Codeint 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 approachimport java.io.*;class GFG{ // Function to calculate the timestatic 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 Codepublic 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 approachusing System;class GFG{ // Function to calculate the timestatic 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 Codepublic 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 timefunction 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 timefunction 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 Codevar 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> |
8 Hours
Time Complexity: O(1)
Auxiliary Space: O(1)
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