Description
https://leetcode.com/problems/maximum-nesting-depth-of-the-parentheses/
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(Aconcatenated withB), whereAandBare VPS‘s, or
- It can be written as (A), whereAis a VPS.
We can similarly define the nesting depth depth(S) of any VPS S as follows:
- depth("") = 0
- depth(C) = 0, where- Cis a string with a single character not equal to- "("or- ")".
- depth(A + B) = max(depth(A), depth(B)), where- Aand- Bare VPS‘s.
- depth("(" + A + ")") = 1 + depth(A), where- Ais 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
- sconsists of digits- 0-9and characters- '+',- '-',- '*',- '/',- '(', and- ')'.
- It is guaranteed that parentheses expression sis a VPS.
Explanation
Use a stack to track the left parenthesis encountered, if encountering a right parenthesis, just pop the stack. Return the max depth of the stack.
Python Solution
class Solution:
    def maxDepth(self, s: str) -> int:
        
        stack = []
        
        max_depth = 0
        for c in s:
            if c == '(':
                stack.append(c)                
                max_depth = max(max_depth, len(stack))
            elif c == ')':
                stack.pop()
                
        return max_depth
                - Time Complexity: O(N).
- Space Complexity: O(N).