"""
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

Constraints:
* 1 <= s.length <= 100
* s consists of digits 0-9 and characters '+', '-', '*', '/', '(', and ')'.
* It is guaranteed that parentheses expression s is a VPS.


Solution:
Count open brackets.
Ignore numbers and signs.
Only brackets matter. Only count the current open brackets.
Encounter left bracket, depth + 1; Encounter right bracket, depth -1.
The max depth equals to the max open bracket number.
"""


# Count open brackets
# TC: O(N), N = len(s)
# SC: O(1)
class Solution:
    def maxDepth(self, s: str) -> int:
        max_depth, cur = 0, 0
        for w in s:
            if w == '(':
                cur += 1
            elif w == ')':
                cur -= 1
            max_depth = max(max_depth, cur)
        return max_depth
