LeetCode-in-Java

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 = 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.

Solution

import java.util.ArrayList;
import java.util.List;

public class Solution {
    public String removeOuterParentheses(String s) {
        List<String> primitives = new ArrayList<>();
        int i = 1;
        while (i < s.length()) {
            int initialI = i - 1;
            int left = 1;
            while (i < s.length() && left > 0) {
                if (s.charAt(i) == '(') {
                    left++;
                } else {
                    left--;
                }
                i++;
            }
            primitives.add(s.substring(initialI, i));
            i++;
        }
        StringBuilder sb = new StringBuilder();
        for (String primitive : primitives) {
            sb.append(primitive, 1, primitive.length() - 1);
        }
        return sb.toString();
    }
}

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