# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` `

Output :

```5
```

Time Complexity : O(N/X)

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