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

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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 <bits/stdc++.h>
using namespace std;
 
// Driver code
int main(){
 
    int n = 7, k = 3;
 
    if(n % k == 0)
        cout<<n/k<<" "<<n/k;
    else
        cout<<((n-(n % k))/k)<<" "
            <<(((n-(n % k))/k) + 1);
 
    return 0;
}
 
// This code is contributed by Sanjit_Prasad

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

<?php
// PHP implementation of the above approach
 
// Driver code
$n = 7; $k = 3;
 
if($n % $k == 0)
    echo $n/$k . " " . $n/$k;
else
    echo (($n - ($n % $k)) / $k) . " " .
        ((($n - ($n % $k)) / $k) + 1);
 
// This code is contributed 
// by Akanksha Rai(Abby_akku)
?>

Javascript

<script>
 
// JavaScript implementation of the above approach 
 
 
// Driver code
var n = 7 ;
var k = 3 ;
 
if (n % k == 0)
    document.write(n / k +" " + n / k);
 
else
    document.write((n-(n % k)) / k + " "
            + (((n-(n % k))/ k) + 1) ) ;
 
 
// This code is contributed by 29AjayKumar 
 
</script>

Output:

2 3

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

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