1614. 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

  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 (Java)

class Solution {
    public int maxDepth(String s) {
        Deque<Character> stack = new ArrayDeque<>();
        int maxDepth = 0;
        for (char c : s.toCharArray()) {
            if (c == '(') {
                stack.push(c);
            } else if (c == ')') {
                maxDepth = Math.max(maxDepth, stack.size());
                stack.pop();
            }
        }
        return maxDepth;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).