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856. Score of Parentheses

Description

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.

Solutions

• Java
• C++
• Python
• Go

• class Solution {
      public int scoreOfParentheses(String s) {
          int ans = 0, d = 0;
          for (int i = 0; i < s.length(); ++i) {
              if (s.charAt(i) == '(') {
                  ++d;
              } else {
                  --d;
                  if (s.charAt(i - 1) == '(') {
                      ans += 1 << d;
                  }
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int scoreOfParentheses(string s) {
          int ans = 0, d = 0;
          for (int i = 0; i < s.size(); ++i) {
              if (s[i] == '(') {
                  ++d;
              } else {
                  --d;
                  if (s[i - 1] == '(') {
                      ans += 1 << d;
                  }
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def scoreOfParentheses(self, s: str) -> int:
          ans = d = 0
          for i, c in enumerate(s):
              if c == '(':
                  d += 1
              else:
                  d -= 1
                  if s[i - 1] == '(':
                      ans += 1 << d
          return ans
  
  

• func scoreOfParentheses(s string) int {
  	ans, d := 0, 0
  	for i, c := range s {
  		if c == '(' {
  			d++
  		} else {
  			d--
  			if s[i-1] == '(' {
  				ans += 1 << d
  			}
  		}
  	}
  	return ans
  }
  

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