makes sense ty

This solution has a time complexity of O(n^2) in the worst case scenario, where n is the length of the input string.

This is due to the use of the contains() method, which loops through the remaining substring to check if a character appears more than once.

In the worst case scenario, this will cause the loop to be repeated n times, resulting in O(n^2) time complexity.

chic!

I don't get it. You iterate through the array only once. What if opposite direction pairs appear after you perform some other deletions?

is it be more better if we use {for}?

Stack is a legacy class that should not be used in new code. So this is not best practice. (Deque should be used instead of Stack)
https://stackoverflow.com/questions/37314298/why-stack-extends-vector-in-jdk

yes, i have same opinion, using api to simplify the code but it have poor performance

Doesn't matter. In the end it's only important whether the number of 'w's and 'e's was equal. Which one is +1 and which one is -1 doesn't change that.

I think case 'w' and 'e' are exchanged!
case 'w' should be x--;
case 'e' should be x++;

Kind of embarassing that a O(n^2) is this high up :/

Nice job!