1614

1614. Maximum Nesting Depth of the Parentheses (Easy)

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(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.

Related Topics:
String

Similar Questions:

• Maximum Nesting Depth of Two Valid Parentheses Strings (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/maximum-nesting-depth-of-the-parentheses/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maxDepth(string s) {
        int ans = 0, left = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] != '(' && s[i] != ')') continue;
            if (s[i] == '(') ++left;
            else --left;
            ans = max(ans, left);
        }
        return ans;
    }
};