Remove Invalid Parentheses

                                            
class Solution {
public:
    vector<string> removeInvalidParentheses(string s) {
        
        vector<string> result;
        unordered_set<string> visited;
        queue<string> q;
        q.push(s);
        visited.insert(s);
        bool found = false;
        
        while (!q.empty()) {
            string current = q.front();
            q.pop();
            
            if (isValid(current)) {
                result.push_back(current);
                found = true;
            }
            
            if (found) continue;
            
            for (int i = 0; i < current.length(); i++) {
                if (current[i] != '(' && current[i] != ')') continue;
                
                string next = current.substr(0, i) + current.substr(i + 1);
                
                if (visited.find(next) == visited.end()) {
                    q.push(next);
                    visited.insert(next);
                }
            }
        }
        
        return result;
    }
    
    bool isValid(string s) {
        int count = 0;
        for (char c : s) {
            if (c == '(') count++;
            else if (c == ')') {
                if (count == 0) return false;
                count--;
            }
        }
        return count == 0;
    }
};

                                            
class Solution {
    public List<String> removeInvalidParentheses(String s) {
        List<String> result = new ArrayList<>();
        if (s == null) {
            return result;
        }
        
        Set<String> visited = new HashSet<>();
        Queue<String> queue = new LinkedList<>();
        boolean found = false;
        
        queue.offer(s);
        visited.add(s);
        
        while (!queue.isEmpty()) {
            s = queue.poll();
            
            if (isValid(s)) {
                result.add(s);
                found = true;
            }
            
            if (found) {
                continue;
            }
            
            for (int i = 0; i < s.length(); i++) {
                if (s.charAt(i) != '(' && s.charAt(i) != ')') {
                    continue;
                }
                
                String newStr = s.substring(0, i) + s.substring(i + 1);
                
                if (!visited.contains(newStr)) {
                    queue.offer(newStr);
                    visited.add(newStr);
                }
            }
        }
        
        return result;
    }
    
    private boolean isValid(String s) {
        int count = 0;
        
        for (char c : s.toCharArray()) {
            if (c == '(') {
                count++;
            } else if (c == ')') {
                count--;
                if (count < 0) {
                    return false;
                }
            }
        }
        
        return count == 0;
    }
}

                                            
class Solution(object):
    def removeInvalidParentheses(self, s):
        def isValid(s):
            cnt = 0
            for char in s:
                if char == '(':
                    cnt += 1
                elif char == ')':
                    cnt -= 1
                    if cnt < 0:
                        return False
            return cnt == 0
        
        def dfs(s, start, left, right):
            if left == 0 and right == 0:
                if isValid(s):
                    res.append(s)
                return
            
            for i in range(start, len(s)):
                if i > start and s[i] == s[i-1]:
                    continue
                
                if right > 0 and s[i] == ')':
                    dfs(s[:i] + s[i+1:], i, left, right-1)
                
                elif left > 0 and s[i] == '(':
                    dfs(s[:i] + s[i+1:], i, left-1, right)
        
        left = 0
        right = 0
        for char in s:
            if char == '(':
                left += 1
            elif char == ')':
                if left > 0:
                    left -= 1
                else:
                    right += 1
        
        res = []
        dfs(s, 0, left, right)
        return res

                                            
public class Solution {
    public IList<string> RemoveInvalidParentheses(string s) {
        List<string> result = new List<string>();
        HashSet<string> visited = new HashSet<string>();
        Queue<string> queue = new Queue<string>();
        bool found = false;
        
        queue.Enqueue(s);
        visited.Add(s);
        
        while(queue.Count > 0 && !found){
            int size = queue.Count;
            for(int i = 0; i < size; i++){
                string current = queue.Dequeue();
                
                if(IsValid(current)){
                    result.Add(current);
                    found = true;
                }
                
                if(!found){
                    for(int j = 0; j < current.Length; j++){
                        if(current[j] != '(' && current[j] != ')') continue;
                        
                        string next = current.Substring(0, j) + current.Substring(j + 1);
                        
                        if(!visited.Contains(next)){
                            visited.Add(next);
                            queue.Enqueue(next);
                        }
                    }
                }
            }
        }
        
        return result;
    }
    
    private bool IsValid(string s){
        int count = 0;
        
        foreach(char c in s){
            if(c == '('){
                count++;
            } else if(c == ')'){
                count--;
                if(count < 0) return false;
            }
        }
        
        return count == 0;
    }
}

                                            
function removeInvalidParentheses(s: string): string[] {
    const result: string[] = [];

    const isValid = (str: string): boolean => {
        let count = 0;
        for (let char of str) {
            if (char === '(') {
                count++;
            } else if (char === ')') {
                count--;
                if (count < 0) return false;
            }
        }
        return count === 0;
    }

    const dfs = (str: string, start: number, open: string, close: string): void => {
        let count = 0;
        for (let i = start; i < str.length; i++) {
            if (str[i] === open) count++;
            if (str[i] === close) count--;
            if (count < 0) {
                for (let j = 0; j <= i; j++) {
                    if (str[j] === close && (j === 0 || str[j - 1] !== close)) {
                        dfs(str.substring(0, j) + str.substring(j + 1), i, open, close);
                    }
                }
                return;
            }
        }

        const reversed = str.split('').reverse().join('');
        if (open === '(') {
            dfs(reversed, 0, close, open);
        } else {
            if (isValid(reversed) && !result.includes(reversed)) {
                result.push(reversed);
            }
        }
    }

    dfs(s, 0, '(', ')');

    return result;
};

                                            
class Solution {
    /**
     * @param String $s
     * @return String[]
     */
    function removeInvalidParentheses($s) {
        $result = [];
        
        $this->dfs($s, 0, 0, ['(', ')'], $result);
        
        return $result;
    }
    
    function dfs($s, $start, $lastToRemove, $par, &$result) {
        $count = 0;
        for ($i = $start; $i < strlen($s); $i++) {
            if ($s[$i] == $par[0]) {
                $count++;
            } elseif ($s[$i] == $par[1]) {
                $count--;
            }
            
            if ($count < 0) {
                for ($j = $lastToRemove; $j <= $i; $j++) {
                    if ($s[$j] == $par[1] && ($j == $lastToRemove || $s[$j] != $s[$j - 1])) {
                        $newStr = substr($s, 0, $j) . substr($s, $j + 1);
                        $this->dfs($newStr, $i, $j, $par, $result);
                    }
                }
                
                return;
            }
        }
        
        $reversed = strrev($s);
        
        if ($par[0] == '(') {
            $this->dfs($reversed, 0, 0, [')', '('], $result);
        } else {
            $result[] = $reversed;
        }
    }
}

                                            
class Solution {
  List<String> removeInvalidParentheses(String s) {
    List<String> result = [];

    void dfs(int index, int leftCount, int rightCount, int leftRemoval, int rightRemoval, String current) {
      if (index == s.length) {
        if (leftCount == rightCount) {
          if (leftRemoval == 0 && rightRemoval == 0) {
            result.add(current);
          }
        }
        return;
      }

      if (s[index] == '(') {
        if (leftRemoval > 0) {
          dfs(index + 1, leftCount, rightCount, leftRemoval - 1, rightRemoval, current);
        }
        dfs(index + 1, leftCount + 1, rightCount, leftRemoval, rightRemoval, current + '(');
      } else if (s[index] == ')') {
        if (rightRemoval > 0) {
          dfs(index + 1, leftCount, rightCount, leftRemoval, rightRemoval - 1, current);
        }
        if (leftCount > rightCount) {
          dfs(index + 1, leftCount, rightCount + 1, leftRemoval, rightRemoval, current + ')');
        }
      } else {
        dfs(index + 1, leftCount, rightCount, leftRemoval, rightRemoval, current + s[index]);
      }
    }

    int extraLeft = 0;
    int extraRight = 0;

    for (int i = 0; i < s.length; i++) {
      if (s[i] == '(') {
        extraLeft++;
      } else if (s[i] == ')') {
        if (extraLeft > 0) {
          extraLeft--;
        } else {
          extraRight++;
        }
      }
    }

    dfs(0, 0, 0, extraLeft, extraRight, '');

    return result.toSet().toList();
  }
}

To solve this coding challenge, we need to remove the minimum number of invalid parentheses to make the input string valid. The challenge requires us to return all possible valid strings that can be formed by removing the fewest number of parentheses. The procedure involves checking every possible way to remove parentheses and validating them to ensure they result in valid strings.

# Explanation

The given problem mentions that the input string can contain both letters and parentheses, while the parentheses can be misplaced. The valid strings are those where every opening parenthesis "(" has a corresponding closing parenthesis ")", and no closing parenthesis ")" comes before its matching opening one. The algorithm can be broken down into the following detailed steps:

Validation Function : First, we define a helper function that checks if a given string has valid parentheses. This function will iterate through the string, keeping track of the number of open and closed parentheses. If the count of closing parentheses exceeds the opening ones at any point, the string is invalid. Also, by the end of the string, the number of opening and closing parentheses must be balanced.
Depth-First Search (DFS) for Removal : Next, we use DFS to iterate through all possible strings generated by removing a parenthesis at every position. This function keeps track of the starting index for removal and the counts of extra left "(" and right ")" parentheses to be removed. If we reach a point where no extra parentheses need to be removed, we validate the resultant string and add it to the results if valid.
Counting Miscellaneous Parentheses : We first count the extra left and right parentheses which do not have a matching pair. These counts help the DFS function to make targeted removals and avoid unnecessary work.
Storing Valid Results : We maintain a list to store all unique valid strings that can be produced by the minimum removal of parentheses. This list is returned as the result.

Let's delineate the pseudocode clearly, explaining each step:

Detailed Steps in Pseudocode

1. Initializing the Parameters and Data Structures

First, we need to count the mismatched parentheses and initialize the results list.

                                            
define function count_invalid_parentheses(s):
    left_excess <- 0
    right_excess <- 0
    for char in s:
        if char == '(': 
            left_excess += 1
        else if char == ')':
            if left_excess > 0:
                left_excess -= 1
            else:
                right_excess += 1
    return left_excess, right_excess

left_excess, right_excess = count_invalid_parentheses(s)
results = empty list

2. Validity Check Function

This function will check if a string

s

is valid with balanced parentheses.

                                            
define function is_valid(s):
    balance_count <- 0
    for char in s:
        if char == '(':
            balance_count += 1
        else if char == ')':
            balance_count -= 1
            if balance_count < 0:
                return False
    return balance_count == 0

3. DFS Function to Explore All Possible Valid Strings

This function checks all possible removals starting from

start_index

and processes only when there are extra parentheses to be removed.

                                            
define function dfs(s, start_index, left_removal, right_removal, results):
    if left_removal == 0 and right_removal == 0:
        if is_valid(s):
            add s to results
        return

    for i in range from start_index to length of s:
        // Avoid duplicate work
        if i > start_index and s[i] == s[i - 1]:
            continue

        // Recursively remove right parenthesis
        if right_removal > 0 and s[i] == ')':
            dfs(s[0:i] + s[i+1:], i, left_removal, right_removal - 1, results)

        // Recursively remove left parenthesis
        if left_removal > 0 and s[i] == '(':
            dfs(s[0:i] + s[i+1:], i, left_removal - 1, right_removal, results)

4. Call DFS to Generate the Result

Initiate the DFS search starting from the first position in the string.

                                            
dfs(s, 0, left_excess, right_excess, results)
return results

By following the above steps, we ensure that we explore all potential valid strings formed by removing the minimal number of parentheses and return all unique valid strings. This exhaustive procedure ensures that we efficiently and effectively arrive at the correct result using concepts of DFS, validation checks, and careful handling of duplicate processing.