Algorithm to Compute the 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(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.

Hint:
The depth of any character in the VPS is the ( number of left brackets before it ) – ( number of right brackets before it )

Compute the Maximum Nesting Depth of the Parentheses

Similar to most Parentheses string problems. We can use a balance counter to record the current depth. If it is an opening Parentheses, we increment the counter/depth, if it is closing Parentheses, we decrement it accordingly. Then, we just need to get the maximum counter during the process until we reach the end of the valid Parentheses string.

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class Solution {
public:
    int maxDepth(string s) {
        int ans = 0, bal = 0;
        for (const auto &n: s) {
            if (n == '(') {
                ans = max(ans, ++ bal);
            } else if (n == ')') {
                bal --;
            }
        }
        return ans;
    }
};

class Solution {
public:
    int maxDepth(string s) {
        int ans = 0, bal = 0;
        for (const auto &n: s) {
            if (n == '(') {
                ans = max(ans, ++ bal);
            } else if (n == ')') {
                bal --;
            }
        }
        return ans;
    }
};

The time complexity of this algorithm is O(N) where N is the number in the valid parentheses string. We are using the constant space i.e. O(1).

–EOF (The Ultimate Computing & Technology Blog) —