这个题的Time/Space complexities怎么算啊？

LC 301. Remove Invalid Parentheses

      solution： 0ms 100.00%, 9.3MB 95.83%

class Solution {
public:
    vector<string> removeInvalidParentheses(string s) {
        /// recursion forward (')') and backward ('(')

        vector<string> res;
        remove_invalid(s, 0, 0, {'(', ')'}, res);
        return res;
    }

    void remove_invalid(const string& s, int last_idx, int last_rem,
                        const vector<char>& bracket, vector<string>& res) {
        int cnt = 0; // count of the first bracket

        for (int i = last_idx; i < s.size(); ++i) {
            if (s[i] == bracket[0]) ++cnt;
            else if (s[i] == bracket[1]) --cnt;
            
            if (cnt >= 0) continue;  // no obvious violation yet

            // bracket[1] violates, remove
            for (int j = last_rem; j <= i; ++j) {
                if (s[j] == bracket[1] && (j == last_rem || s[j] != s[j-1])) {
                    remove_invalid(s.substr(0, j) + s.substr(j+1), i, j, bracket, res);
                }
            }
            // all done. don't need to go back.
            return;
        }
  
        // up to here meaning need to remove extra left brackets
        // A trick is used to reverse the string and treat '(' as ')'.
        string s_rev = string(s.rbegin(), s.rend());
        if (bracket[0] == '(') remove_invalid(s_rev, 0, 0, { ')', '(' }, res);
        else res.push_back(s_rev); // save result (must've been reversed twice already)
    }
};
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Space 应该是O（N）吧（for string s_rev)？
Time呢？O（2 * N^2) ??