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:

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

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()) {
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();
}
}