[Leet Code] Maximum Nesting Depth of the Parentheses

LeetCode: https://leetcode.com/problems/maximum-nesting-depth-of-the-parentheses/

Problem:

A string is a valid parentheses string (denoted VPS) if it meets one of the following:

• It is an empty string "", or a single character not equal to "(" or ")",
• It can be written as AB (A concatenated with B), where A and B are VPS's, or
• It can be written as (A), where A is a VPS.

We can similarly define the nesting depth depth(S) of any VPS S as follows:

• depth("") = 0
• depth(C) = 0, where C is a string with a single character not equal to "(" or ")".
• depth(A + B) = max(depth(A), depth(B)), where A and B are VPS's.
• depth("(" + A + ")") = 1 + depth(A), where A is a VPS.

For example, "", "()()", and "()(()())" are VPS's (with nesting depths 0, 1, and 2), and ")(" and "(()" are not VPS's.

Given a VPS represented as string s, return the nesting depth of s.

Example 1:

Input: s = "(1+(2*3)+((8)/4))+1"
Output: 3
Explanation: Digit 8 is inside of 3 nested parentheses in the string.

Example 2:

Input: s = "(1)+((2))+(((3)))"
Output: 3

Example 3:

Input: s = "1+(2*3)/(2-1)"
Output: 1

Example 4:

Input: s = "1"
Output: 0

Solution:

class Solution(object):
    def maxDepth(self, s):
        """
        :type s: str
        :rtype: int
        """
        stack = []
        max_val = 0
        for i in s:
            if i == "(":
                stack.append(i)
            if i == ")":
                stack.pop()
            length = len(stack)
            if length > max_val:
                max_val = length
        return max_val

Explanation:

For this problem, we want to create a stack that will hold the amount of "(" we have in our string. If we encounter a "(" we will append it to the stack, otherwise, we will pop it. This will mean that outermost parenthesis will persist if we have something like "(a(b) + cd(e)) the (b) will be removed but it will also increase our stack length, and therefore add to our max val!