I have String s1 and String s2, the pair s1 , s2 is called special strings if we can make them anagram by removing a single character (any number of times) from each string s1 and s2.

I have String s1 and String s2, the pair s1 , s2 is called special strings if we can make them anagram by removing a single character (any number of times) from each string s1 and s2.

I am trying to remove unncessary parentheses in an expression. So far, my code handles most of the use cases such as,

I have a string s of length n made of only 3 characters 0, 1, and !, and two numbers provided as x and y

I have an array of numbers of size n, I want to count the segments, (which are subarrays of size 3 or more) based on below condition:

This question is asked during an online interview:

I came across this problem.

Working on one problem, I’ve two arrays, they are like a grid with N cols with 2 rows

Given integer arrays A and B of size n

Given two arrays a and b of length n. I need to find amount of pairs (l, r), 1 <= l <= r <= n such that max(a[l], ..., a[r]) = min(b[l], ..., b[r]).