Removing Parentheses

https://binarysearch.com/problems/Removing-Parentheses

I know it's not Leetcode please don't kill me. I was in a competition and I thought this was neat. Couldn't find the leetcode alternative.

Given a string of parentheses s, return the minimum number of parentheses to be removed to make the string valid (i.e. each open parenthesis is eventually closed).

For example, given the string "()())()", you should return 1. Given the string ")(", you should return 2, since we must remove all of them.

class Solution:
    def solve(self, s):
        """
        first thought is we track ( and )
        if we one is higher than the other, we remove it
        """
        if not s:
            return 0
        total = 0
        temp = 0
        for p in s:
            if p == "(":
                total += 1
            elif p == ")" and total:
                # the and total means this only runs if total is more than 0
                # so we only have ) if we have ( before it
                total -= 1
            else:
                temp += 1
        return total + temp

We keep track of 2 variables:

Total (will be 0 if we have the same number of ( and ).
Temp - will increment if we ) but no ( before it.

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Last updated 5 years ago