LeetCode 1614. Maximum Nesting Depth of the Parentheses

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 (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.

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).

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