Software Engineer

You are given an infinite supply of coins of each of denominations D = {D0, D1, D2, D3, ...... Dn-1}. You need to figure out the total number of ways W, in which you can make a change for value V using coins of denominations from D. Print 0, if a change isn't possible.

You are given an array consisting of 'N' elements and you need to perform 'Q' queries on the given array. Each query consists of an integer which tells the number of elements by which you need to left rotate the given array. For each query return the final array obtained after performing the left rotations.

Note:

Perform each query on the original array only i.e. every output should be according to the original order of elements.

Example:

Let the array be [1, 2, 3, 4, 5, 6] and the queries be {2, 4, 1}. For every query, we’ll perform the required number of left rotations on the array.

For the first query, rotate the given array to the left by 2 elements, so the resultant array is: [3, 4, 5, 6, 1, 2].

For the second query, rotate the given array to the left by 4 elements, so the resultant array is: [5, 6, 1, 2, 3, 4].

For the third query, rotate the given array to the left by 1 element, so the resultant array is: [2, 3, 4, 5, 6, 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.

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