Associate Cloud Engineer

Given an unsorted array ‘arr’ of distinct integers and an integer ‘k’, your task is to find the ‘k-th’ smallest element in the array.

Example:

n = 5, k = 2 and arr[] = {6, 5, 4, 8, 7}
The array elements in sorted order are [4, 5, 6, 7, 8]. The ‘2-nd’ smallest element in the array is 5, so the answer is 5.

Note:

1. Don’t print anything. Return the value of ‘k-th’ smallest element.
2. ‘k’ is a positive integer and not greater than the size of the array.
3. The array ‘arr’ is unsorted, and all the elements of the array are distinct.

You are given a city which contains 'N' x 'M' houses, where each house can have one of the following three conditions :

1. The value ‘0’ represents an empty house,
2. The value ‘1’ represents a non-infected person,
3. The value ‘2’ represents an infected person.

It takes one day to propagate the infection from an infected house to its adjacent (Front, Back, Left, Right) non-empty and non-infected house. An empty house can only break the line of propagation of infection.

You need to return the minimum number of days Covid will take to infect each and every house in the city. And for the God’s sake if this is impossible, return -1 instead.

You're given two sorted arrays 'arr1' and 'arr2' of size 'n' and 'm' respectively and an element 'k'.

Find the element that would be at the 'kth' position of the combined sorted array.

Position 'k' is given according to 1 - based indexing, but arrays 'arr1' and 'arr2' are using 0 - based indexing.

For example :

Input: 'arr1' = [2, 3, 45], 'arr2' = [4, 6, 7, 8] and 'k' = 4
Output: 6
Explanation: The merged array will be [2, 3, 4, 6, 7, 8, 45]. The element at position '4' of this array is 6. Hence we return 6.

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