MTS 1

You are given an array of integers 'ARR' of length 'N' and an integer Target. Your task is to return all pairs of elements such that they add up to Target.

Note:

We cannot use the element at a given index twice.

Follow Up:

Try to do this problem in O(N) time complexity. 

You have been given a binary matrix of size 'N' * 'M' where each element is either 0 or 1. You are also given a source and a destination cell, both of them lie within the matrix.

Your task is to find the length of the shortest path from the source cell to the destination cell only consisting of 1s. If there is no path from source to destination cell, return -1.

Note:

1. Coordinates of the cells are given in 0-based indexing.
2. You can move in 4 directions (Up, Down, Left, Right) from a cell.
3. The length of the path is the number of 1s lying in the path.
4. The source cell is always filled with 1.

For example -

1 0 1
1 1 1
1 1 1
For the given binary matrix and source cell(0,0) and destination cell(0,2). Few valid paths consisting of only 1s are

X 0 X     X 0 X 
X X X     X 1 X 
1 1 1     X X X 
The length of the shortest path is 5.

Let’s define the beauty of an array to be equal to the value present in the array, which is having the maximum frequency. If there are multiple such values, then the smallest among such values will be selected.

Alice gave Bob an array ‘A’ of size ‘N’ and asked him to find the beauty of the frequencies of the values present in the array. In simple words, Bob first needs to find the frequencies of all values present in the array, then find the beauty of this frequency array.

Help Bob in finding the beauty of the frequencies of the values present in the array.

You are given a matrix 'MAT' of size ‘N’ * ‘M’, where ‘N’ is the number of rows and ‘M’ is the number of columns. Value ‘0’ in a cell represents that the cell is set on fire initially, (at time t = ‘0’), and the cells which don’t have fire initially have value = ‘1’, and are called safe cells.

If a cell is on fire, then in one second the fire will expand to its adjacent cells (left, right, top, bottom) if they are not already on fire.

You are given the position of a person as integers ‘X’ and ‘Y’ denoting the cell (‘X’, ‘Y’). In one second the person can move from its current cell to any one of its adjacent cells, provided they are not on fire.

You have to determine if the person can move through the matrix to one of the escape cells without burning himself i.e. without going through the fire cells. If it’s possible, return time taken, else return -1.

Note:

1. Escape cells in this problem are all such cells that lie on the edge of the matrix but not on the corner. i.e all such cells which are in the first row, first column, last row, and last column excluding the four corner cells are considered as valid escape cells.

2. A cell once set on fire continues to remain on fire till the end.

3. Note that rows are indexed from 0 to ‘N’ - 1 and columns are indexed from 0 to ‘M’ - 1.

4. The escape cells may also be initially on fire or can catch fire from some neighboring cell.

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