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Formatted question description: https://leetcode.ca/all/1021.html

1021. Remove Outermost Parentheses (Easy)

A valid parentheses string is either empty (""), "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation.  For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.

A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.

Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings.

Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.

 

Example 1:

Input: "(()())(())"
Output: "()()()"
Explanation: 
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".

Example 2:

Input: "(()())(())(()(()))"
Output: "()()()()(())"
Explanation: 
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".

Example 3:

Input: "()()"
Output: ""
Explanation: 
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".

 

Note:

1. S.length <= 10000
2. S[i] is "(" or ")"
3. S is a valid parentheses string

 

Companies:
Google

Related Topics:
Stack

Solution 1.

// OJ: https://leetcode.com/problems/remove-outermost-parentheses/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    string removeOuterParentheses(string S) {
        string ans;
        int cnt = 0, start = 0;
        for (int i = 0; i < S.size(); ++i) {
            if (S[i] == '(') ++cnt;
            else if (--cnt == 0) {
                ans += S.substr(start + 1, i - start - 1);
                start = i + 1;
            }
        }
        return ans;
    }
};

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public String removeOuterParentheses(String S) {
          List<Integer> indices = new ArrayList<Integer>();
          Stack<Character> stack = new Stack<Character>();
          int length = S.length();
          int begin = 0;
          for (int i = 0; i < length; i++) {
              char c = S.charAt(i);
              if (c == '(')
                  stack.push(c);
              else {
                  stack.pop();
                  if (stack.isEmpty()) {
                      indices.add(begin);
                      indices.add(i);
                      begin = i + 1;
                  }
              }
          }
          StringBuffer sb = new StringBuffer();
          for (int i = 0; i < length; i++) {
              if (!indices.contains(i))
                  sb.append(S.charAt(i));
          }
          return sb.toString();
      }
  }
  
  ############
  
  class Solution {
      public String removeOuterParentheses(String s) {
          StringBuilder ans = new StringBuilder();
          int cnt = 0;
          for (int i = 0; i < s.length(); ++i) {
              char c = s.charAt(i);
              if (c == '(') {
                  if (++cnt > 1) {
                      ans.append(c);
                  }
              } else {
                  if (--cnt > 0) {
                      ans.append(c);
                  }
              }
          }
          return ans.toString();
      }
  }
  

• // OJ: https://leetcode.com/problems/remove-outermost-parentheses/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      string removeOuterParentheses(string S) {
          string ans;
          int cnt = 0, start = 0;
          for (int i = 0; i < S.size(); ++i) {
              if (S[i] == '(') ++cnt;
              else if (--cnt == 0) {
                  ans += S.substr(start + 1, i - start - 1);
                  start = i + 1;
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def removeOuterParentheses(self, s: str) -> str:
          ans = []
          cnt = 0
          for c in s:
              if c == '(':
                  cnt += 1
                  if cnt > 1:
                      ans.append(c)
              else:
                  cnt -= 1
                  if cnt > 0:
                      ans.append(c)
          return ''.join(ans)
  
  ############
  
  class Solution(object):
      def maxScoreSightseeingPair(self, A):
          """
          :type A: List[int]
          :rtype: int
          """
          pre_i = 0
          res = 0
          for j in range(1, len(A)):
              res = max(A[j] - j + A[pre_i] + pre_i, res)
              if A[j] + j > A[pre_i] + pre_i:
                  pre_i = j
          return res
  

• func removeOuterParentheses(s string) string {
  	ans := []rune{}
  	cnt := 0
  	for _, c := range s {
  		if c == '(' {
  			cnt++
  			if cnt > 1 {
  				ans = append(ans, c)
  			}
  		} else {
  			cnt--
  			if cnt > 0 {
  				ans = append(ans, c)
  			}
  		}
  	}
  	return string(ans)
  }
  

• function removeOuterParentheses(s: string): string {
      let res = '';
      let depth = 0;
      for (const c of s) {
          if (c === '(') {
              depth++;
          }
          if (depth !== 1) {
              res += c;
          }
          if (c === ')') {
              depth--;
          }
      }
      return res;
  }
  
  

• impl Solution {
      pub fn remove_outer_parentheses(s: String) -> String {
          let mut res = String::new();
          let mut depth = 0;
          for c in s.chars() {
              if c == '(' {
                  depth += 1;
              }
              if depth != 1 {
                  res.push(c);
              }
              if c == ')' {
                  depth -= 1;
              }
          }
          res
      }
  }
  
  

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