SDE - 1

You are given a two-dimensional matrix of integers of dimensions N*M, where each cell represents the number of coins in that cell. Alice and Bob have to collect the maximum number of coins. The followings are the conditions to collect coins:

Alice starts from top left corner, i.e., (0, 0) and should reach left bottom corner, i.e., (N-1, 0). Bob starts from top right corner, i.e., (0, M-1) and should reach bottom right corner, i.e., (N-1, M-1).

From a point (i, j), Alice and Bob can move to (i+1, j+1) or (i+1, j-1) or (i+1, j)

They have to collect all the coins that are present at a cell. If Alice has already collected coins of a cell, then Bob gets no coins if goes through that cell again.

For example :

If the matrix is 
0 2 4 1
4 8 3 7
2 3 6 2
9 7 8 3
1 5 9 4

Then answer is 47. As, Alice will collect coins 0+8+3+9+1 = 21 coins. Bob will collect coins 1+7+6+8+4 = 26 coins. Total coins is 21+26 = 47 coins.

You have been given an array/list 'WORDS' of 'N' non-empty words, and an integer 'K'. Your task is to return the 'K' most frequent words sorted by their frequency from highest to lowest.

Note:

If two words have the same frequency then the lexicographically smallest word should come first in your answer.

Follow up:

Can you solve it in O(N * logK) time and O(N) extra space? 

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 have been given a binary matrix 'MAT' containing only 0’s and 1’s of size N x M. You need to find the distance of the nearest cell having 1 in the matrix for each cell.

The distance is calculated as |i1 – i2| + |j1 – j2|, where i1, j1 are the coordinates of the current cell and i2, j2 are the coordinates of the nearest cell having value 1.

Note :

You can only move in four directions which are : Up, Down, Left and Right.

For example :

If N = 3, M = 4

and mat[ ][ ] = { 0, 0, 0, 1,
                  0, 0, 1, 1,
                  0, 1, 1, 0 }

then the output matrix will be

3  2  1  0
2  1  0  0
1  0  0  1

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