LeetCode 1021. Remove Outermost Parentheses Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1

PythonJavaC++GoTypeScriptRust

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)(code-box)

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(); } }(code-box)

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; } };(code-box)

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) }(code-box)

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; }(code-box)

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 } }(code-box)

Solution 2

PythonJavaC++Go

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) if c == ')': cnt -= 1 return ''.join(ans)(code-box)

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 == '(') { ++cnt; } if (cnt > 1) { ans.append(c); } if (c == ')') { --cnt; } } return ans.toString(); } }(code-box)

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

func removeOuterParentheses(s string) string { ans := []rune{} cnt := 0 for _, c := range s { if c == '(' { cnt++ } if cnt > 1 { ans = append(ans, c) } if c == ')' { cnt-- } } return string(ans) }(code-box)