# Minimum and maximum number of N chocolates after distribution among K students

Last Updated : 18 Jun, 2022

Given N Chocolates and K students, the task is to find how to divide the chocolates such that the difference between the minimum and maximum chocolate received by all students is minimized. Print the value of minimum and maximum chocolate distribution.
Examples

```Input: N = 7, K = 3
Output: Min = 2, Max = 3
Distribution is 2 2 3

Input: N = 100, K = 10
Output: 10 10
Distribution is 10 10 10 10 10 10 10 10 10 10 ```

Approach: The difference will only be minimized when each student gets an equal number of candies that is N % k = 0 but if N % K != 0 then each student will 1st get (N-N%k)/k amount of candy then the rest N%k amount of candies can be distributed to N%K students by giving them each 1 candy. Thus there will be just 1 more candy than the (N-N%k)/k if N % K != 0 with a student.
Below is the implementation of the above approach:

## CPP

 `// CPP implementation of the above approach` `#include ` `using` `namespace` `std;`   `// Driver code` `int` `main(){`   `    ``int` `n = 7, k = 3;`   `    ``if``(n % k == 0)` `        ``cout<

## Java

 `// Java implementation of the above approach `   `public` `class` `Improve {` `    `  `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `            ``int` `n = ``7` `;` `            ``int` `k = ``3` `;` `            `  `            ``if` `(n % k == ``0``)` `                ``System.out.println(n / k +``" "` `+ n / k);` `            `  `            ``else` `                ``System.out.println((n-(n % k)) / k + ``" "` `                        ``+ (((n-(n % k))/ k) + ``1``) ) ;`   `    ``}` `    ``// This Code is contributed by ANKITRAI1` `}`

## Python

 `# Python implementation of the above approach`   `n, k ``=` `7``, ``3` `if``(n ``%` `k ``=``=` `0``):` `    ``print``(n``/``/``k, n``/``/``k)`   `else``:` `    ``print``((n``-``n ``%` `k)``/``/``k, (n``-``n ``%` `k)``/``/``k ``+` `1``)`

## C#

 `// C# implementation of the ` `// above approach ` `using` `System;`   `class` `GFG ` `{`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``int` `n = 7 ;` `    ``int` `k = 3 ;` `    `  `    ``if` `(n % k == 0)` `        ``Console.WriteLine(n / k + ` `                    ``" "` `+ n / k);` `    `  `    ``else` `        ``Console.WriteLine((n - (n % k)) / k + ` `                  ``" "` `+ (((n - (n % k)) / k) + 1));` `}` `}`   `// This code is contributed ` `// by inder_verama`

## PHP

 ``

## Javascript

 ``

Output:

`2 3`

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

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