301. Remove Invalid Parentheses

Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.

Note: The input string may contain letters other than the parentheses ( and ).

Example 1:

Input: "()())()"
Output: ["()()()", "(())()"]

Example 2:

Input: "(a)())()"
Output: ["(a)()()", "(a())()"]

Example 3:

Input: ")("
Output: [""]

 

Approach #1: C++.

class Solution {
public:
    vector<string> removeInvalidParentheses(string s) {
        int l = 0;
        int r = 0;
　　　　 // calculate the mismatch parentheses.
        for (char c : s) {
            l += (c == '(');
            if (l == 0) {
                r += (c == ')');
            } else {
                l -= (c == ')');
            }
        }
        
        vector<string> ans;
        
        dfs(s, 0, l, r, ans);
        
        return ans;
    }
private:
    bool isValid(string& s) {
        int count = 0;
        for (char c : s) {
            if (c == '(') count++;
            if (c == ')') count--;
            if (count < 0) return false;
        }
        return count == 0;
    }
    
    void dfs(string s, int start, int l, int r, vector<string>& ans) {
        if (l == 0 && r == 0) {
            if (isValid(s))　　// judge the string is valid.
                ans.push_back(s);
            return;
        }
        
        for (int i = start; i < s.length(); ++i) {
            if (i != start && s[i] == s[i-1]) continue;
　　　　　　　 // s[i] is '(' or ')'
            if (s[i] == '(' || s[i] == ')') {
                string curr = s;
                curr.erase(i, 1);
                if (r > 0) dfs(curr, i, l, r-1, ans);
                else if (l > 0) dfs(curr, i, l-1, r, ans);
            }
        }
    }
};

　　

Analysis:

step1: compute min number of '(' and ')' to remove, unbalanced ')' + unblanced '('

step2: try all possible ways to remove r '(' and l ')'. Remove '(' first to make prefix valid.

step3: when r == 0 and l == 0, judging the string is or not fulfiled conditions.