Minimum number of equal amount bags to collect at least M money

Given unlimited number of coins of two denomination X and Y. Also given bags with capacity of N rupees, independent of number of coins. The task is to find the minimum number of bags such that each bag contains the same amount of rupees and sum of all the bags amount is at least M.

Examples :

Input : M = 27, N = 12, X = 2, Y = 5. 
Output : 3
We put 2 coins of X, 1 coin of Y in each bag. 
So we have 9 rupees in each bag and we need 
at least 3 bags (Note that 27/9 = 3). There 
is no way to obtain sum with lesser number
of bags.

Input : M = 45, N = 9, X = 4, Y = 5. 
Output : 5

The task is to minimize the number of bags, thus need to maximize the amount in a bag such that amount in all bags is same. Suppose we take, ‘p’ number of X coins and ‘q’ number of Y coins, then the task is to maximize p*X + q*Y. And also, p*X + q*Y <= N.
Now, to find the maximum possible value of Left Hand Side of the equation, vary p from 0 to N/X, then find the maximum possible q for the particular p. Then, out of all such pairs, take the (p, q) pair which gives maximum value of p*X + q*Y. 

Below is the implementation of above idea : 

C++

// C++ program to find minimum number of bags such
// that each bag contains same amount and sum is at
// least M.
#include<bits/stdc++.h>
using namespace std;

// Return minimum number of bags required.
int minBags(int M, int N, int X, int Y)
{
    // Initialize maximum amount in a bag
    int maxAmount = 0;

    // Finding maximum possible q for the particular p.
    for (int p = 0; p <= N/X; p++)
    {
        int q = (N - p * X) / Y;

        maxAmount = max(maxAmount, p*X + q*Y);
    }

    // Calculating the minimum number of bags.
    int result = M/maxAmount;
    result += (M % maxAmount == 0? 0: 1);

    return result;
}

// Driven Program
int main()
{
    int M = 45, N = 9;
    int X = 4, Y = 5;

    cout << minBags(M, N, X, Y) << endl;

    return 0;
}

Java

// Java program to find minimum number
// of bags such that each bag contains
// same amount and sum is at least M
import java.io.*;

public class GFG {

// Return minimum number of bags required.
static int minBags(int M, int N,
                   int X, int Y)
{

    // Initialize maximum amount in a bag
    int maxAmount = 0;

    // Finding maximum possible q
    // for the particular p.
    for (int p = 0; p <= N / X; p++)
    {
        int q = (N - p * X) / Y;

        maxAmount = Math.max(maxAmount, p * X +
                                        q * Y);
    }

    // Calculating the minimum number of bags.
    int result = M / maxAmount;
    result += (M % maxAmount == 0? 0: 1);

    return result;
}

    // Driver Code
    static public void main (String[] args)
    {
        int M = 45, N = 9;
        int X = 4, Y = 5;

        System.out.println(minBags(M, N, X, Y));
    }
}

// This code is contributed by vt_m.

Python3

# Python 3 program to find minimum number
# of bags such that each bag contains same
# amount and sum is at least M.

# Return minimum number of bags required.
def minBags(M, N, X, Y):

    # Initialize maximum amount in a bag
    maxAmount = 0

    # Finding maximum possible q for
    # the particular p.
    for p in range(0, int(N / X) + 1, 1):
        q = int((N - p * X) / Y)

        maxAmount = max(maxAmount, p * X + q * Y)

    # Calculating the minimum number of bags.
    result = int(M / maxAmount)
    if(M % maxAmount == 0):
        result += 0
    else:
        result += 1

    return result

# Driver Code
if __name__ == '__main__':
    M = 45
    N = 9
    X = 4
    Y = 5

    print(minBags(M, N, X, Y))

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find minimum number of
// bags such that each bag contains same
// amount and sum is at least M.
using System;

public class GFG
{

// Return minimum number of bags required.
static int minBags(int M, int N,
                int X, int Y)
{

    // Initialize maximum amount in a bag
    int maxAmount = 0;

    // Finding maximum possible q
    // for the particular p.
    for (int p = 0; p <= N / X; p++)
    {
        int q = (N - p * X) / Y;

        maxAmount = Math.Max(maxAmount, p * X +
                                        q * Y);
    }

    // Calculating the minimum number of bags.
    int result = M / maxAmount;
    result += (M % maxAmount == 0? 0: 1);

    return result;
}

    // Driver Code
    static public void Main ()
    {
        int M = 45, N = 9;
        int X = 4, Y = 5;

        Console.WriteLine(minBags(M, N, X, Y));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find minimum
// number of bags such that each
// bag contains same amount and
// sum is at least M.

// Return minimum number
// of bags required.
function minBags($M, $N, $X, $Y)
{
    // Initialize maximum
    // amount in a bag
    $maxAmount = 0;

    // Finding maximum possible
    // q for the particular p.
    for ($p = 0; $p <= $N / $X; $p++)
    {
        $q = ($N - $p * $X) / $Y;

        $maxAmount = max($maxAmount,
                         $p * $X + $q * $Y);
    }

    // Calculating the minimum
    // number of bags.
    $result = $M / $maxAmount;
    $result += ($M % $maxAmount == 0? 0: 1);

    return $result;
}

// Driver Code
$M = 45; $N = 9;
$X = 4 ; $Y = 5;

echo minBags($M, $N, $X, $Y) ;

// This code is contributed by nitin mittal.
?>

Javascript

<script>
// Javascript program to find minimum number
// of bags such that each bag contains
// same amount and sum is at least M

// Return minimum number of bags required.
function minBags(M, N, X, Y)
{

    // Initialize maximum amount in a bag
    let maxAmount = 0;

    // Finding maximum possible q
    // for the particular p.
    for (let p = 0; p <= N / X; p++)
    {
        let q = (N - p * X) / Y;

        maxAmount = Math.max(maxAmount, p * X +
                                        q * Y);
    }

    // Calculating the minimum number of bags.
    let result = M / maxAmount;
    result += (M % maxAmount == 0? 0: 1);

    return result;
}

// Driver code

    let M = 45, N = 9;
       let X = 4, Y = 5;

       document.write(minBags(M, N, X, Y));

      // This code is contributed by susmitakundugoaldanga.
</script>

Output :

5

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