856. Score of Parentheses

Problem

Tags: String, Stack

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.

Code

C

// 856. Score of Parentheses (9/24/54176)
// Runtime: 0 ms (93.55%) Memory: 5.50 MB (70.97%) 

int scoreOfParentheses (char * s) {
    int score = 0, depth = 0;
    
    for (int i = 0; s[i] != '\0'; i++) {
        if (s[i] == '(') {
            if (s[i + 1] == ')') {
                score += 1 << depth;
                i++;
            } else {
                depth++;
            }
        } else {
            depth--;
        }
    }
    
    return score;
}

GO

// 856. Score of Parentheses (10/4/54176)
// Runtime: 0 ms (82.14%) Memory: 1.90 MB (0.00%) 

func scoreOfParentheses(s string) int {
    score, depth := 0, 0
    
    for i := 0; i < len(s); i++ {
        if s[i] == '(' {
            if s[i + 1] != s[i] {
                score += 1 << depth
                i++
            } else {
                depth++
            }
        } else {
            depth--
        }
    }
    
    return score
}

TS

// 856. Score of Parentheses (9/28/54176)
// Runtime: 72 ms (63.64%) Memory: 42.82 MB (54.55%) 

function scoreOfParentheses(s: string): number {
    let score = 0, depth = 0;
    
    for (let i = 0; i < s.length; i++) {
        if (s[i] === "(") {
            if (s[i + 1] !== s[i]) {
                score += 1 << depth;
                i++;
            } else {
                depth++;
            }
        } else {
            depth--;
        }
    }
    
    return score;
};