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Maximize count of empty water bottles from N filled bottles
  • Difficulty Level : Basic
  • Last Updated : 10 Mar, 2021

Given two integers N and E, where N represents the number of full water bottles and E represents the number of empty bottles that can be exchanged for a full water bottle. The task is to find the maximum number of water bottles that can be emptied. 

Examples:

Input: N = 9, E = 3
Output: 13
Explanation:
Initially, there are 9 fully filled water bottles. 
All of them are emptied to obtain 9 empty bottles. Therefore, count = 9
Then exchange 3 bottles at a time for 1 fully filled bottle. 
Therefore, 3 fully filled bottles are obtained. 
Then those 3 bottles are emptied to get 3 empty bottles. Therefore, count = 9 + 3 = 12
Then those 3 bottles are exchanged for 1 full bottle which is emptied. 
Therefore, count = 9 + 3 + 1 = 13.

Input: N = 7, E = 5
Output: 8
Explanation:
Empty the 7 fully filled water bottles. Therefore, count = 7
Then exchange 5 bottles to obtain 1 fully filled bottle. Then empty that bottle. 
Therefore, count = 7 + 1 = 8.

Approach: The following steps have to be followed to solve the problem:



  • Store the value of N in a temporary variable, say a. 
  • Initialize a variable, say s, to store the count of empty bottles and another variable, say b, to store the count of bottles remaining after exchange.
  • Now, iterate until a is not equal to 0 and increment s  by a, as it will be the minimum number of bottles that have been emptied.
  • Update the following values:
    • a to (a + b) / e
    • b to N – (a * e)
    • N to (a+b)
  • Return s as the required answer.

Below is the implementation of the above approach:

C++




// C++ program for the
// above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum
// bottles that can be emptied
int maxBottles(int n, int e)
{
    int s = 0, b = 0;
    int a = n;
 
    // Iterate until a
  // is non-zero
    while (a != 0) {
       
        // Add the number of
      // bottles that are emptied
        s = s + a;
 
        // Update a after
        // exchanging empty bottles
        a = (a + b) / e;
 
        // Stores the number of bottles
        // left after the exchange
        b = n - (a * e);
        n = a + b;
    }
 
    // Return the answer
    return s;
}
 
// Driver Code
int main()
{
    int n = 9, e = 3;
 
    // Function call
    int s = maxBottles(n, e);
    cout << s << endl;
}

Java




// Java program for the
// above approach
import java.util.*;
 
class GFG{
 
// Function to find the maximum
// bottles that can be emptied
static int maxBottles(int n, int e)
{
    int s = 0, b = 0;
    int a = n;
 
    // Iterate until a
    // is non-zero
    while (a != 0)
    {
     
        // Add the number of
        // bottles that are emptied
        s = s + a;
 
        // Update a after
        // exchanging empty bottles
        a = (a + b) / e;
 
        // Stores the number of bottles
        // left after the exchange
        b = n - (a * e);
        n = a + b;
    }
 
    // Return the answer
    return s;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 9, e = 3;
 
    // Function call
    int s = maxBottles(n, e);
     
    System.out.print(s + "\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program for the
# above approach
 
# Function to find the maximum
# bottles that can be emptied
def maxBottles(n, e):
     
    s = 0
    b = 0
    a = n
 
    # Iterate until a
    # is non-zero
    while (a != 0):
 
        # Add the number of
        # bottles that are emptied
        s = s + a
 
        # Update a after
        # exchanging empty bottles
        a = (a + b) // e
 
        # Stores the number of bottles
        # left after the exchange
        b = n - (a * e)
        n = a + b
 
    # Return the answer
    return s
 
# Driver Code
if __name__ == '__main__':
     
    n = 9
    e = 3
 
    # Function call
    s = maxBottles(n, e)
    print(s)
 
# This code is contributed by mohit kumar 29

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the maximum
// bottles that can be emptied
static int maxBottles(int n, int e)
{
    int s = 0, b = 0;
    int a = n;
 
    // Iterate until a
    // is non-zero
    while (a != 0)
    {
     
        // Add the number of
        // bottles that are emptied
        s = s + a;
 
        // Update a after
        // exchanging empty bottles
        a = (a + b) / e;
 
        // Stores the number of bottles
        // left after the exchange
        b = n - (a * e);
        n = a + b;
    }
 
    // Return the answer
    return s;
}
 
// Driver Code
public static void Main()
{
    int n = 9, e = 3;
 
    // Function call
    int s = maxBottles(n, e);
     
    Console.Write(s + "\n");
}
}
 
// This code is contributed by code_hunt

Javascript




<script>
 
// javascript program for the
// above approach
 
// Function to find the maximum
// bottles that can be emptied
function maxBottles(n, e)
{
    var s = 0, b = 0;
    var a = n;
 
    // Iterate until a
    // is non-zero
    while (a != 0)
    {
     
        // Add the number of
        // bottles that are emptied
        s = s + a;
 
        // Update a after
        // exchanging empty bottles
        a = (a + b) / e;
 
        // Stores the number of bottles
        // left after the exchange
        b = n - (a * e);
        n = a + b;
    }
 
    // Return the answer
    return s;
}
 
// Driver Code
 
var n = 9, e = 3;
 
// Function call
var s = maxBottles(n, e);
 
document.write(parseInt(s) + "\n");
 
// This code contributed by Princi Singh
 
</script>

Output:

13

Time Complexity: O(N)
Auxiliary Space: O(1)

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