Remove Invalid Parentheses

An expression will be given which can contain open and close parentheses and optionally some characters, No other operator will be there in string. We need to remove minimum number of parentheses to make the input string valid. If more than one valid output are possible removing same number of parentheses then print all such output. 
Examples:

Input  : str = “()())()” -
Output : ()()() (())()
There are two possible solutions
"()()()" and "(())()"

Input  : str = (v)())()
Output : (v)()()  (v())()

As we need to generate all possible output we will backtrack among all states by removing one opening or closing bracket and check if they are valid if invalid then add the removed bracket back and go for next state. We will use BFS for moving through states, use of BFS will assure removal of minimal number of brackets because we traverse into states level by level and each level corresponds to one extra bracket removal. Other than this BFS involve no recursion so overhead of passing parameters is also saved. 
Below code has a method isValidString to check validity of string, it counts open and closed parenthesis at each index ignoring non-parenthesis character. If at any instant count of close parenthesis becomes more than open then we return false else we keep update the count variable. 

C++

/*  C/C++ program to remove invalid parenthesis */
#include <bits/stdc++.h>
using namespace std;

//  method checks if character is parenthesis(open
// or closed)
bool isParenthesis(char c)
{
    return ((c == '(') || (c == ')'));
}

//  method returns true if string contains valid
// parenthesis
bool isValidString(string str)
{
    int cnt = 0;
    for (int i = 0; i < str.length(); i++)
    {
        if (str[i] == '(')
            cnt++;
        else if (str[i] == ')')
            cnt--;
        if (cnt < 0)
            return false;
    }
    return (cnt == 0);
}

//  method to remove invalid parenthesis
void removeInvalidParenthesis(string str)
{
    if (str.empty())
        return ;

    //  visit set to ignore already visited string
    unordered_set<string> visit;

    //  queue to maintain BFS
    queue<string> q;
    string temp;
    bool level;

    //  pushing given string as starting node into queue
    q.push(str);
    visit.insert(str);
    while (!q.empty())
    {
        str = q.front();  q.pop();
        if (isValidString(str))
        {
            cout << str << endl;

            // If answer is found, make level true
            // so that valid string of only that level
            // are processed.
            level = true;
        }
        if (level)
            continue;
        for (int i = 0; i < str.length(); i++)
        {
            if (!isParenthesis(str[i]))
                continue;

            // Removing parenthesis from str and
            // pushing into queue,if not visited already
            temp = str.substr(0, i) + str.substr(i + 1);
            if (visit.find(temp) == visit.end())
            {
                q.push(temp);
                visit.insert(temp);
            }
        }
    }
}

//  Driver code to check above methods
int main()
{
    string expression = "()())()";
    removeInvalidParenthesis(expression);

    expression = "()v)";
    removeInvalidParenthesis(expression);

    return 0;
}

Java

// Java program to remove invalid parenthesis
import java.util.*;

class GFG
{

// method checks if character is parenthesis(open
// or closed)
static boolean isParenthesis(char c)
{
    return ((c == '(') || (c == ')'));
}

// method returns true if string contains valid
// parenthesis
static boolean isValidString(String str)
{
    int cnt = 0;
    for (int i = 0; i < str.length(); i++)
    {
        if (str.charAt(i) == '(')
            cnt++;
        else if (str.charAt(i) == ')')
            cnt--;
        if (cnt < 0)
            return false;
    }
    return (cnt == 0);
}

// method to remove invalid parenthesis
static void removeInvalidParenthesis(String str)
{
    if (str.isEmpty())
        return;

    // visit set to ignore already visited string
    HashSet<String> visit = new HashSet<String>();

    // queue to maintain BFS
    Queue<String> q = new LinkedList<>();
    String temp;
    boolean level = false;

    // pushing given string as
    // starting node into queue
    q.add(str);
    visit.add(str);
    while (!q.isEmpty())
    {
        str = q.peek(); q.remove();
        if (isValidString(str))
        {
            System.out.println(str);

            // If answer is found, make level true
            // so that valid string of only that level
            // are processed.
            level = true;
        }
        if (level)
            continue;
        for (int i = 0; i < str.length(); i++)
        {
            if (!isParenthesis(str.charAt(i)))
                continue;

            // Removing parenthesis from str and
            // pushing into queue,if not visited already
            temp = str.substring(0, i) + str.substring(i + 1);
            if (!visit.contains(temp))
            {
                q.add(temp);
                visit.add(temp);
            }
        }
    }
}

// Driver Code
public static void main(String[] args)
{
    String expression = "()())()";
    removeInvalidParenthesis(expression);

    expression = "()v)";
    removeInvalidParenthesis(expression);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to remove invalid parenthesis

# Method checks if character is parenthesis(open
# or closed)
def isParenthesis(c):
    return ((c == '(') or (c == ')'))

# method returns true if contains valid
# parenthesis
def isValidString(str):
    cnt = 0
    for i in range(len(str)):
        if (str[i] == '('):
            cnt += 1
        else if (str[i] == ')'):
            cnt -= 1
        if (cnt < 0):
            return False
    return (cnt == 0)

# method to remove invalid parenthesis
def removeInvalidParenthesis(str):
    if (len(str) == 0):
        return

    # visit set to ignore already visited
    visit = set()

    # queue to maintain BFS
    q = []
    temp = 0
    level = 0

    # pushing given as starting node into queue
    q.append(str)
    visit.add(str)
    while(len(q)):
        str = q[0]
        q.pop()
        if (isValidString(str)):
            print(str)

            # If answer is found, make level true
            # so that valid of only that level
            # are processed.
            level = True
        if (level):
            continue
        for i in range(len(str)):
            if (not isParenthesis(str[i])):
                continue

            # Removing parenthesis from str and
            # pushing into queue,if not visited already
            temp = str[0:i] + str[i + 1:]
            if temp not in visit:
                q.append(temp)
                visit.add(temp)

# Driver Code
expression = "()())()"
removeInvalidParenthesis(expression)
expression = "()v)"
removeInvalidParenthesis(expression)

# This code is contributed by SHUBHAMSINGH10

C#

// C# program to remove invalid parenthesis
using System;
using System.Collections.Generic;

class GFG
{

// method checks if character is
// parenthesis(open or closed)
static bool isParenthesis(char c)
{
    return ((c == '(') || (c == ')'));
}

// method returns true if string contains
// valid parenthesis
static bool isValidString(String str)
{
    int cnt = 0;
    for (int i = 0; i < str.Length; i++)
    {
        if (str[i] == '(')
            cnt++;
        else if (str[i] == ')')
            cnt--;
        if (cnt < 0)
            return false;
    }
    return (cnt == 0);
}

// method to remove invalid parenthesis
static void removeInvalidParenthesis(String str)
{
    if (str == null || str == "")
        return;

    // visit set to ignore already visited string
    HashSet<String> visit = new HashSet<String>();

    // queue to maintain BFS
    Queue<String> q = new Queue<String>();
    String temp;
    bool level = false;

    // pushing given string as
    // starting node into queue
    q.Enqueue(str);
    visit.Add(str);
    while (q.Count != 0)
    {
        str = q.Peek(); q.Dequeue();
        if (isValidString(str))
        {
            Console.WriteLine(str);

            // If answer is found, make level true
            // so that valid string of only that level
            // are processed.
            level = true;
        }

        if (level)
            continue;
        for (int i = 0; i < str.Length; i++)
        {
            if (!isParenthesis(str[i]))
                continue;

            // Removing parenthesis from str and
            // pushing into queue,if not visited already
            temp = str.Substring(0, i) +
                   str.Substring(i + 1);
            if (!visit.Contains(temp))
            {
                q.Enqueue(temp);
                visit.Add(temp);
            }
        }
    }
}

// Driver Code
public static void Main(String[] args)
{
    String expression = "()())()";
    removeInvalidParenthesis(expression);

    expression = "()v)";
    removeInvalidParenthesis(expression);
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// JavaScript program to remove invalid parenthesis

// method checks if character is parenthesis(open
// or closed)
function isParenthesis(c)
{
    return ((c == '(') || (c == ')'));
}

// method returns true if string contains valid
// parenthesis
function isValidString(str)
{
    let cnt = 0;
    for (let i = 0; i < str.length; i++)
    {
        if (str[i] == '(')
            cnt++;
        else if (str[i] == ')')
            cnt--;
        if (cnt < 0)
            return false;
    }
    return (cnt == 0);
}

// method to remove invalid parenthesis
function removeInvalidParenthesis(str)
{
    if (str.length==0)
        return;

    // visit set to ignore already visited string
    let visit = new Set();

    // queue to maintain BFS
    let q = [];
    let temp;
    let level = false;

    // pushing given string as
    // starting node into queue
    q.push(str);
    visit.add(str);
    while (q.length!=0)
    {
        str = q.shift();
        if (isValidString(str))
        {
            document.write(str+"<br>");

            // If answer is found, make level true
            // so that valid string of only that level
            // are processed.
            level = true;
        }
        if (level)
            continue;
        for (let i = 0; i < str.length; i++)
        {
            if (!isParenthesis(str[i]))
                continue;

            // Removing parenthesis from str and
            // pushing into queue,if not visited already
            temp = str.substring(0, i) + str.substring(i + 1);
            if (!visit.has(temp))
            {
                q.push(temp);
                visit.add(temp);
            }
        }
    }
}

// Driver Code
let expression = "()())()";
removeInvalidParenthesis(expression);

expression = "()v)";
removeInvalidParenthesis(expression);

// This code is contributed by rag2127

</script>

Output:

(())()
()()()
(v)
()v

The time and space complexity of the given program are as follows:

Time complexity:
The program uses a BFS (breadth-first search) approach to traverse all possible combinations of removing invalid parentheses from the input string. The worst-case time complexity of BFS is exponential, O(b^d), where b is the branching factor (the average number of child nodes per node) and d is the depth of the search tree. In the given program, the branching factor is at most 2 (either remove or keep a parenthesis at each position), and the depth can be up to the length of the input string. Therefore, the worst-case time complexity of the program is O(2^n), where n is the length of the input string.

Space complexity:
The program uses a queue to store the intermediate strings during the BFS traversal. The maximum size of the queue can be at most O(2^n), as there can be up to 2^n valid combinations of parentheses. Additionally, the program uses an unordered set to keep track of the visited strings and avoid duplicates. The size of the set can also be up to O(2^n), as each valid combination can potentially be a unique string. Therefore, the space complexity of the program is O(2^n).

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