SDE - Intern

You are given two strings BEGIN and END and an array of strings DICT. Your task is to find the length of the shortest transformation sequence from BEGIN to END such that in every transformation you can change exactly one alphabet and the word formed after each transformation must exist in DICT.

Note:

1. If there is no possible path to change BEGIN to END then just return -1.
2. All the words have the same length and contain only lowercase english alphabets.
3. The beginning word i.e. BEGIN will always be different from the end word i.e. END (BEGIN != END).

You are given an array containing ‘N’ words. For each word, you need to find its shortest prefix which can uniquely identify it. For example “abcd” and “abdc” both have the prefix “ab” in common so we can’t uniquely find a word using the prefix “ab”. To uniquely identify both the words we need the prefix “abc” from “abcd” and “abd” from “abdc”.

Note:

You can assume that the words are unique. It means that it is always possible to find a unique prefix for each word.

Given a ‘N’ * ’M’ maze with obstacles, count and return the number of unique paths to reach the right-bottom cell from the top-left cell. A cell in the given maze has a value '-1' if it is a blockage or dead-end, else 0. From a given cell, we are allowed to move to cells (i+1, j) and (i, j+1) only. Since the answer can be large, print it modulo 10^9 + 7.

For Example :

Consider the maze below :
0 0 0 
0 -1 0 
0 0 0

There are two ways to reach the bottom left corner - 

(1, 1) -> (1, 2) -> (1, 3) -> (2, 3) -> (3, 3)
(1, 1) -> (2, 1) -> (3, 1) -> (3, 2) -> (3, 3)

Hence the answer for the above test case is 2.

You are given a 'Nx3' 2-D array 'Jobs' describing 'N' jobs where 'Jobs[i][0]' denotes the id of 'i-th' job, 'Jobs[i][1]' denotes the deadline of 'i-th' job, and 'Jobs[i][2]' denotes the profit associated with 'i-th job'.

You will make a particular profit if you complete the job within the deadline associated with it. Each job takes 1 unit of time to be completed, and you can schedule only one job at a particular time.

Return the number of jobs to be done to get maximum profit.

Note :

If a particular job has a deadline 'x', it means that it needs to be completed at any time before 'x'.

Assume that the start time is 0.

For Example :

'N' = 3, Jobs = [[1, 1, 30], [2, 3, 40], [3, 2, 10]].

All the jobs have different deadlines. So we can complete all the jobs.

At time 0-1, Job 1 will complete.
At time 1-2, Job 3 will complete.
At time 2-3, Job 2 will complete.

So our answer is [3 80].

You are given an array ‘ARR’ consist of ‘N’ positive integers, and a positive integer ‘K’.

Your task is to count and return the number of contiguous subarrays having products of their elements strictly less than ‘K’.

Example:

Consider an array ARR = [1, 2, 3, 4, 5], and K = 7, then all the subarrays having product less than 7 are  [1], [2], [3], [4], [5], [1, 2], [2,3], [1,2,3]   i.e there is total 8 subarays having product less than 7.

A thief is robbing a store and can carry a maximal weight of W into his knapsack. There are N items and the ith item weighs wi and is of value vi. Considering the constraints of the maximum weight that a knapsack can carry, you have to find and return the maximum value that a thief can generate by stealing items.

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.