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 codeint 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 approachn, k = 7, 3if(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 codepublic 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 codevar 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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