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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.

  • 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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