Maximum nesting depth of parentheses in leetcode simple questions

subject

If the string meets one of the following conditions , It can be called Valid bracket string （valid parentheses string, I could just write it as VPS）：
The string is an empty string “”, Or a not for “(” or “)” The single character of .
The string can be written as AB（A And B String connection ）, among A and B All are Valid bracket string .
The string can be written as (A), among A It's a Valid bracket string .
Similarly , You can define any valid parenthesis string S Of Nesting depth depth(S)：
depth("") = 0
depth = 0, among C Is a single character string , And the character is not “(” perhaps “)”
depth(A + B) = max(depth(A), depth(B)), among A and B All are Valid bracket string
depth("(" + A + “)”) = 1 + depth(A), among A It's a Valid bracket string
for example ：""、"()()"、"()(()())" All are Valid bracket string （ The nesting depth is 0、1、2）, and “)(” 、"(()" Are not Valid bracket string .
To give you one Valid bracket string s, Returns the of the string s Nesting depth .
Example 1：
Input ：s = “(1+(23)+((8)/4))+1"
Output ：3
explain ： Numbers 8 In the nested 3 Layer in parentheses .
Example 2：
Input ：s = “(1)+((2))+(((3)))”
Output ：3
Example 3：
Input ：s = "1+(23)/(2-1)”
Output ：1
Example 4：
Input ：s = “1”
Output ：0
Tips ：
1 <= s.length <= 100
s By digital 0-9 And character ‘+’、’-’、’*’、’/’、’(’、’)’ form
Title Data guarantee bracket expression s yes Valid parenthesis expression
source ： Power button （LeetCode）

Their thinking

   Traversing the entire string , Add parentheses on one side only ’(‘ or ’)' And update the maximum depth . In addition, because the given string in the title has been guaranteed to be a valid kueha expression , So we don't need to update the maximum depth every time we accumulate parentheses , The maximum depth can be updated only when the opposite bracket is encountered .

class Solution:
    def maxDepth(self, s: str) -> int:
        MAX=0
        count=0
        for i in s:
            if i=='(':
                count+=1
            elif i==')':
                if MAX<count:
                    MAX=count
                count-=1
        return MAX