##### 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:

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;
}
};

• 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();
}
}

############

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
}
}