301. Remove Invalid Parentheses

Posted 2020-11-08 The Tech Road

class Solution {
public:
    unordered_set<string> res;    // DFS may produce dup result, use set to filter.
    vector<string> removeInvalidParentheses(string s) {
        // Calculate rmL and rmR for the numbers of () we need to remove.
        int rmL = 0, rmR = 0;
        for (auto c : s) {
            if (c == ‘(‘)   rmL++;
            else if (c == ‘)‘) {
                if (rmL > 0)    rmL--;
                else    rmR++;
            }
        }
        dfs(s, "", 0, rmL, rmR, 0);
        return vector<string>(res.begin(), res.end());
    }
    void dfs(const string& s, string path, int cur, int rmL, int rmR, int open) {
        // Ignore non()
        while (cur < s.length() && s[cur] != ‘(‘ && s[cur] != ‘)‘) {
            path.push_back(s[cur]);
            cur++;
        }
        if (cur == s.length()) {
            if (open == 0 && rmL == 0 && rmR == 0) {
                res.insert(path);
            }
            return;
        }
        if (s[cur] == ‘(‘) {
            if (rmL > 0)    // we can remove (, and we remove.
                dfs(s, path, cur+1, rmL-1, rmR, open);
            dfs(s, path + "(", cur+1, rmL, rmR, open+1);    // we keep this (.
        }
        else if (s[cur] == ‘)‘) {
            if (rmR > 0)    // we can remove ), and we remove.
                dfs(s, path, cur+1, rmL, rmR-1, open);
            if (open > 0)   // we have ( to match, keep it.
                dfs(s, path+")", cur+1, rmL, rmR, open-1);
        }
    }
};