856. Score of Parentheses

Problem

Given a balanced parentheses string s, return **the *score* of the string**.

The score of a balanced parentheses string is based on the following rule:

• "()" has score 1.

• AB has score A + B, where A and B are balanced parentheses strings.

• (A) has score 2 * A, where A is a balanced parentheses string.

  Example 1:

Input: s = "()"
Output: 1

Example 2:

Input: s = "(())"
Output: 2

Example 3:

Input: s = "()()"
Output: 2

  Constraints:

• 2 <= s.length <= 50

• s consists of only '(' and ')'.

• s is a balanced parentheses string.

Solution (Java)

class Solution {
    public int scoreOfParentheses(String s) {
        Deque<Integer> stack = new ArrayDeque<>();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(') {
                stack.push(-1);
            } else {
                int curr = 0;
                while (stack.peek() != -1) {
                    curr += stack.pop();
                }
                stack.pop();
                stack.push(curr == 0 ? 1 : curr * 2);
            }
        }
        int score = 0;
        while (!stack.isEmpty()) {
            score += stack.pop();
        }
        return score;
    }
}

Explain:

nope.

Complexity:

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