Welcome to Subscribe On Youtube

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.

  • 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
    }
}

All Problems

All Solutions