SDE - 1

Given an array/list of length ‘n’, where the array/list represents the boards and each element of the given array/list represents the length of each board. Some ‘k’ numbers of painters are available to paint these boards. Consider that each unit of a board takes 1 unit of time to paint.

You are supposed to return the area of the minimum time to get this job done of painting all the ‘n’ boards under a constraint that any painter will only paint the continuous sections of boards.

Example :

Input: arr = [2, 1, 5, 6, 2, 3], k = 2

Output: 11

Explanation:
First painter can paint boards 1 to 3 in 8 units of time and the second painter can paint boards 4-6 in 11 units of time. Thus both painters will paint all the boards in max(8,11) = 11 units of time. It can be shown that all the boards can't be painted in less than 11 units of time.

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.

You are given two strings 'str1' and 'str1'.

You have to tell whether these strings form an anagram pair or not.

The strings form an anagram pair if the letters of one string can be rearranged to form another string.

Pre-requisites:

Anagrams are defined as words or names that can be formed by rearranging the letters of another word. Such as "spar" can be formed by rearranging letters of "rasp". Hence, "spar" and "rasp" are anagrams. 

Other examples include:

'triangle' and 'integral'
'listen' and 'silent'

Note:

Since it is a binary problem, there is no partial marking. Marks will only be awarded if you get all the test cases correct.