# Chocolate Problem

Moderate
0/80
Average time to solve is 15m
Contributed by

## Problem statement

Given an array/list of integer numbers 'CHOCOLATES' of size 'N', where each value of the array/list represents the number of chocolates in the packet. There are ‘M’ number of students and the task is to distribute the chocolate to their students. Distribute chocolate in such a way that:

1. Each student gets at least one packet of chocolate.

2. The difference between the maximum number of chocolate in a packet and the minimum number of chocolate in a packet given to the students is minimum.

Example :

Given 'N' : 5 (number of packets) and 'M' : 3 (number of students)

And chocolates in each packet is : {8, 11, 7, 15, 2}

All possible way to distribute 5 packets of chocolates among 3 students are -

( 8,15, 7 ) difference of maximum-minimum is ‘15 - 7’ = ‘8’
( 8, 15, 2 ) difference of maximum-minimum is ‘15 - 2’ = ‘13’
( 8, 15, 11 ) difference of maximum-minimum is ‘15 - 8’ = ‘7’
( 8, 7, 2 ) difference of maximum-minimum is ‘8 - 2’ = ‘6’
( 8, 7, 11 ) difference of maximum-minimum is ‘11 - 7’ = ‘4’
( 8, 2, 11 ) difference of maximum-minimum is ‘11 - 2’ = ‘9’
( 15, 7, 2 ) difference of maximum-minimum is ‘15 - 2’ = 13’
( 15, 7, 11 ) difference of maximum-minimum is ‘15 - 7’ = ‘8’
( 15, 2, 11 ) difference of maximum-minimum is ‘15 - 2’ = ‘13’
( 7, 2, 11 ) difference of maximum-minimum is ‘11 - 2’ = ‘9’

Hence there are 10 possible ways to distribute ‘5’ packets of chocolate among the ‘3’ students and difference of combination (8, 7, 11) is ‘maximum - minimum’ = ‘11 - 7’ = ‘4’ is minimum in all of the above.
Detailed explanation ( Input/output format, Notes, Images )
Constraints:
1 <= T <= 50
2 <= M <= N <= 10^4
1 <= CHOCOLATES[i] <= 10^9

Time Limit : 1 sec
2
3 2
7 2 4
4 3
3 5 1 6
2
3
##### Explanation For Sample Input 1 :
Test Case 1 :

All possible way to distribute 3 packets of chocolate among 2 students are -

( 7, 2 ) difference is ‘7 - 2’ = ‘5’
( 7, 4 ) difference is ‘7 - 4’ = ‘3’
( 2, 4 ) difference is ‘4 - 2’ = ‘2’

There are three ways to distribute 3 packets of chocolate among 2 students but pair ( 4, 2 ) has minimum difference in ‘maximum - minimum’ = ‘4 - 2’ = ‘2’

Test Case 2 :

All possible way to distribute 4 packets of chocolate among 3 students are -

( 3, 5, 1 )  difference is ‘5 - 1’ = ‘4’
( 5, 1, 6 )  difference is ‘6 - 1’ = ‘5’
( 1, 6, 3 )  difference is ‘6 - 1’ = ‘5’
( 6, 5, 3 )  difference is  ‘6 - 3’ = ‘3’

There are four options to choose 3 packets of chocolate. Only ( 6, 5, 3 ) pair has the minimum difference ‘6 - 3’ = ‘3’ comparing other pair of difference ( 4, 5, 5 )