# LeetCode 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: [""]
```

## Solution:

[English Version in Youtube](https://youtu.be/hFGjMV851OU)

[中文版解答Youtube Link](https://youtu.be/I91977216Hk)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1i54y1a7g9/)

```
// Solution 1: BFS
class Solution {
    
    int GetErrors(const string &s) {
        stack<char> mystack;
        int valid = 0;
        for (char ch : s) {
            if (ch == '(') {
                mystack.push(ch);
            } else if (ch == ')') {
                if (!mystack.empty()) {
                    valid += 2;
                    mystack.pop();
                }
            } else {
                valid++;
            }
        }
        return s.length() - valid;
    }
    
public:
    vector<string> removeInvalidParentheses(string s) {
        queue<string> myqueue;
        unordered_set<string> visited;
        myqueue.push(s);
        visited.insert(s);
        
        int errors = GetErrors(s);
        while (!myqueue.empty() && errors > 0) {
            int queue_size = myqueue.size();
            for (int k = 0; k < queue_size; k++) {
                string str = myqueue.front();
                myqueue.pop();
                int i = 0;
                while (i < str.length()) {
                    if (str[i] != '(' && str[i] != ')') {
                        i++;
                        continue;
                    }
                    string tmp = str.substr(0,i) + str.substr(i+1);
                    if (visited.count(tmp) == 0) {
                        int current_errors = GetErrors(tmp);
                        if (current_errors < errors) {
                            myqueue.push(tmp);
                            visited.insert(tmp);
                        }
                    }
    
                    while (i+1 < str.length() && str[i] == str[i+1]) i++;
                    i++;
                }
            }
            
            errors--;
        }
        
        vector<string> result;
        while (!myqueue.empty()) {
            string tmp = myqueue.front();
            myqueue.pop();
            result.push_back(tmp);
        }
        
        return result;
    }
};


// Solution 2: DFS
class Solution {
    
    int GetErrors(const string &s) {
        stack<char> mystack;
        int valid = 0;
        for (char ch : s) {
            if (ch == '(') {
                mystack.push(ch);
            } else if (ch == ')') {
                if (!mystack.empty()) {
                    valid += 2;
                    mystack.pop();
                }
            } else {
                valid++;
            }
        }
        return s.length() - valid;
    }
    
public:
    
    void dfs(const string& s, int ind, int validLen, int leftCnt, int rightCnt, string cur, unordered_set<string>& visited, unordered_set<string>& res_set) {
        if (cur.size() == validLen) {
            if (leftCnt == rightCnt) {
                res_set.insert(cur);
            }
            return;
        }
        if (ind >= s.size()) {
            return;
        }
        
        string key = to_string(ind) + cur;
        if (visited.count(key) > 0) return;
        visited.insert(key);
        
        if (s[ind] == '(' || s[ind] == ')') {
            if (cur.length() + (s.size() - ind) > validLen) {
                dfs(s, ind+1, validLen, leftCnt, rightCnt, cur, visited, res_set);
            }
            if (s[ind] == '(') {
                if (leftCnt < validLen / 2) {
                    dfs(s, ind+1, validLen, leftCnt+1, rightCnt, cur + '(', visited, res_set);
                }
            } else if (leftCnt > rightCnt) {
                dfs(s, ind+1, validLen, leftCnt, rightCnt+1, cur + ')', visited, res_set);
            }
        } else {
            dfs(s, ind+1, validLen, leftCnt, rightCnt, cur + s[ind], visited, res_set);
        }
    }
    
    vector<string> removeInvalidParentheses(string s) {
        int validLen = s.size() - GetErrors(s);
        unordered_set<string> visited;
        unordered_set<string> res_set;
        dfs(s, 0, validLen, 0, 0, "", visited, res_set);
        return vector<string>(res_set.begin(), res_set.end());
    }
};

```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://zhenchaogan.gitbook.io/leetcode-solution/leetcode-301-remove-invalid-parentheses.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.