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
#include <bits/stdc++.h>
using namespace std;
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;
}
|
Java
public class Improve {
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 ) ) ;
}
}
|
Python
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#
using System;
class GFG
{
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));
}
}
|
PHP
<?php
$n = 7; $k = 3;
if ( $n % $k == 0)
echo $n / $k . " " . $n / $k ;
else
echo (( $n - ( $n % $k )) / $k ) . " " .
((( $n - ( $n % $k )) / $k ) + 1);
?>
|
Javascript
<script>
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) ) ;
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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