SDE - 2

You are given an Integer array ‘ARR’ and an Integer ‘K’.

Your task is to find the ‘K’ most frequent elements in ‘ARR’. Return the elements in any order.

For Example:

You are given ‘ARR’ = {1, 2, 2, 3, 3} and ‘K’ = 2. 

The answer will {2, 3} as 2 and 3 are the elements occurring most times.

For a given array with N elements, you need to find the length of the longest subsequence from the array such that all the elements of the subsequence are sorted in strictly increasing order.

Strictly Increasing Sequence is when each term in the sequence is larger than the preceding term.

For example:

[1, 2, 3, 4] is a strictly increasing array, while [2, 1, 4, 3] is not.

You have been given a grid containing some oranges. Each cell of this grid has one of the three integers values:

Value 0 - representing an empty cell.

Value 1 - representing a fresh orange.

Value 2 - representing a rotten orange.

Every second, any fresh orange that is adjacent(4-directionally) to a rotten orange becomes rotten.

Your task is to find out the minimum time after which no cell has a fresh orange. If it's impossible to rot all the fresh oranges then print -1.

Note:

1. The grid has 0-based indexing.
2. A rotten orange can affect the adjacent oranges 4 directionally i.e. Up, Down, Left, Right.

You have been given an array 'ARR' of ‘N’ integers. You have to return the minimum number of jumps needed to reach the last index of the array i.e ‘N - 1’.

From index ‘i’, we can jump to an index ‘i + k’ such that 1<= ‘k’ <= ARR[i] .

'ARR[i]' represents the maximum distance you can jump from the current index.

If it is not possible to reach the last index, return -1.

Note:

Consider 0-based indexing.

Example:

Consider the array 1, 2, 3, 4, 5, 6 
We can Jump from index 0 to index 1
Then we jump from index 1 to index 2
Then finally make a jump of 3 to reach index N-1

There is also another path where
We can Jump from index 0 to index 1
Then we jump from index 1 to index 3
Then finally make a jump of 2 to reach index N-1

So multiple paths may exist but we need to return the minimum number of jumps in a path to end which here is 3.

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