1190. Reverse Substrings Between Each Pair of Parentheses

class Solution {
    public String reverseParentheses(String s) {
        Deque<StringBuilder> stack = new ArrayDeque<>();
        StringBuilder current = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (c == '(') {
                stack.push(current);
                current = new StringBuilder();
            } else if (c == ')') {
                current.reverse();
                current = stack.pop().append(current);
            } else {
                current.append(c);
            }
        }
        return current.toString();
    }
}

class Solution:
    def reverseParentheses(self, s: str) -> str:
        stack = []
        current = []

        for c in s:
            if c == '(':
                stack.append(current)
                current = []
            elif c == ')':
                current.reverse()
                current = stack.pop() + current
            else:
                current.append(c)

        return ''.join(current)

class Solution {
public:
    string reverseParentheses(string s) {
        stack<string> st;
        string current = "";

        for (char c : s) {
            if (c == '(') {
                st.push(current);
                current = "";
            } 
            else if (c == ')') {
                reverse(current.begin(), current.end());
                current = st.top() + current;
                st.pop();
            } 
            else {
                current += c;
            }
        }

        return current;
    }
};

class Solution {
    reverseParentheses(s) {
        const stack = [];
        let current = [];

        for (const c of s) {
            if (c === '(') {
                stack.push(current);
                current = [];
            } 
            else if (c === ')') {
                current.reverse();
                current = stack.pop().concat(current);
            } 
            else {
                current.push(c);
            }
        }

        return current.join('');
    }
}