Java Developer

Bob lives with his wife in a city named Berland. Bob is a good husband, so he goes out with his wife every Friday to ‘Arcade’ mall.

‘Arcade’ is a very famous mall in Berland. It has a very unique transportation method between shops. Since the shops in the mall are laying in a straight line, you can jump on a very advanced trampoline from the shop i, and land in any shop between (i) to (i + Arr[i]), where Arr[i] is a constant given for each shop.

There are N shops in the mall, numbered from 0 to N-1. Bob's wife starts her shopping journey from shop 0 and ends it in shop N-1. As the mall is very crowded on Fridays, unfortunately, Bob gets lost from his wife. So he wants to know, what is the minimum number of trampoline jumps from shop 0 he has to make in order to reach shop N-1 and see his wife again. If it is impossible to reach the last shop, return -1.

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.