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1021. Remove Outermost Parentheses

Description

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 = P1 + P2 + ... + Pk, where Pi 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: s = "(()())(())"
Output: "()()()"
Explanation: 
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".

Example 2:

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

Example 3:

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

 

Constraints:

  • 1 <= s.length <= 105
  • s[i] is either '(' or ')'.
  • s is a valid parentheses string.

Solutions

  • 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();
    }
}

  • class Solution {
public:
    string removeOuterParentheses(string s) {
        string ans;
        int cnt = 0;
        for (char& c : s) {
            if (c == '(') {
                if (++cnt > 1) {
                    ans.push_back(c);
                }
            } else {
                if (--cnt) {
                    ans.push_back(c);
                }
            }
        }
        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)


  • 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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