remove-outermost-parentheses.md

Remove Outermost Parentheses

📝 Problem

You’re given a valid parentheses string. A valid string is something like (), (()), or (()()). A string is primitive if it can't be split into two nonempty valid parentheses strings.

Your task is to remove the outermost parentheses from each primitive part of the string and return the result.

📌 Examples

Example 1

Input: s = "(()())(())"
Output: "()()()"

Example 2

Input: s = "(()())(())(()(()))"
Output: "()()()()(())"

Example 3

Input: s = "()()"
Output: ""

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✅ Solutions

💡 Solution 1: method1

function removeOuterParenthesesMethod1 (s) {
    let ans = '';
    let cnt = 0;
    for (const ch of s) {
        if (ch === '(') {
            if (cnt > 0) ans += ch;
            cnt++;
        } else {
            cnt--;
            if (cnt > 0) ans += ch;
        }
    }
    return ans;
};

💡 Solution 2: Two Pointer

function removeOuterParentheses (s) {
    let ans = '';
    let cnt = 0;
    let left = 0;
    
    for (let right = 0; right < s.length; right++) {
        cnt += (s[right] === '(') ? 1 : -1;
        
        if (cnt === 0) {
            ans += s.slice(left + 1, right);
            left = right + 1;
        }
    }
    
    return ans;
};

💡 Solution 3: Recursive Approach

function removeOuterParenthesesRecursion (s) {
    let ans = '';
    let stack = [];
    let i = 0;

    while (i < s.length) {
        if (s[i] === '(') {
            stack.push(s[i]);
            if (stack.length > 1) ans += s[i];
        } else {
            if (stack.length > 1) ans += s[i];
            stack.pop();
        }

        if (stack.length === 0) {
            ans += removeOuterParentheses(s.slice(i + 1));
            break;
        }

        i++;
    }
    
    return ans;
};