Problem Setter Intern

Given a 2D binary matrix of ‘m’ rows and ‘n’ columns initially containing all ‘0’s. The given matrix represents a map where ‘0’s represents water and 1’s represents the land.
You have to perform X operations wherein each operation, you are given a location[i] = (ri, ci) the position at which you should add a land i.e ‘1’.

You need to find the count of islands after performing each of those X operations.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume that the matrix is completely surrounded by water.

1. Frame a complete problem statement with examples, constraints on the inputs, test case format for the input/output, and including sample test cases.

2. Writing all the possible approaches from brute to optimal.

3. Write the solution code for all the approaches in either Java, C++, or Python.

You are given 2 non-empty strings 's1' and 's2' consisting of lowercase English alphabets only.

In one operation you can do either of the following on 's1':

(1) Remove a character from any position in 's1'.

(2) Add a character to any position in 's1'.

Find the minimum number of operations required to convert string 's1' into 's2'.

Example:

Input: 's1' = "abcd", 's2' = "anc"

Output: 3

Explanation:
Here, 's1' = "abcd", 's2' = "anc".
In one operation remove 's1[3]', after this operation 's1' becomes "abc".    
In the second operation remove 's1[1]', after this operation 's1' becomes "ac".
In the third operation add 'n' in 's1[1]', after this operation 's1' becomes "anc".

Hence, the minimum operations required will be 3. It can be shown that there's no way to convert s1 into s2 in less than 3 moves.

You have been given an array/list ‘ARR’ of integers consisting of ‘N’ integers. You are also given an integer ‘X’. In one operation, you can either remove the leftmost or the rightmost element and add that to the sum of removed elements so far. Your task is to return the minimum number of operations such that the sum of removed elements becomes exactly ‘X’. If it is not possible return -1.

For Example :

Let’s say you have an array/list [1, 3, 5, 3] and ‘X’ is 7. 
We can first remove the rightmost element i.e. 3. The array becomes [1,3,5]. In the next step, we can remove the leftmost element i.e 1. The array becomes [3,5] and the sum of removed elements so far becomes 4. In the next step, we can remove the leftmost element i.e 3. The array becomes [5] and the sum of removed elements so far is 7. We have reached our target X i.e 7. Therefore the minimum number of operations to reach ‘X’ is 3.

Tell me about yourself.

What about your coding experience?