Print all Balanced Brackets Strings that can be formed by replacing wild card ‘?’

Given string str containing characters ‘?’, ‘(‘ and ‘)’, the task is to replace the ‘?’ character with ‘(‘ or ‘)’ and print all the strings containing balanced brackets

Example:

Input: str = “????”
Output:
()()
(())

Input: str = “(()?”
Output: (())

Approach: The given problem can be solved using recursion and backtracking. The idea is to substitute every ‘?’ character with ‘)’ then make a recursive call to the next index and after backtracking change it to ‘(‘ then make a recursive call to the next index, after backtracking change the character back to ‘?’. Follow the steps below to solve the problem:

• Convert the string str into a character array, say ch
• Pass the character array ch and index 0 as parameters inside recursive function and at every recursive call perform the following:
  • If the index becomes equal to the length of the character array the:
    • Check if the character array is a balanced bracket string
    • If the above condition is true then print the string
  • If the current character ch[index] is either ‘(‘ or ‘)’ then make a recursive call on the next index
  • If the current character ch[index] is ‘?’ then:
    • Replace it with ‘(‘ and make a recursive call on the next index
    • Replace it with ‘)’ and make a recursive call on the next index
    • Change it back to ‘?’ before returning from the function

Below is the implementation of the above approach:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to print the
// characters of the string
void print(string ch)
{

    for (char c : ch) {
        cout << c;
    }
    cout << endl;
}

// Function to check if the
// brackets are valid or not
bool check(string ch)
{

    // Initialize a stack
    stack<char> S;

    // If character is an open bracket
    // then return false
    if (ch[0] == ')') {

        return false;
    }

    // Iterate the character array
    for (int i = 0; i < ch.length(); i++) {

        // If character is an open bracket
        // then push it in the stack
        if (ch[i] == '(') {

            S.push('(');
        }

        // If character is a close bracket
        else {

            // If stack is empty, there is no
            // corresponding opening bracket
            // so return false
            if (S.size() == 0)
                return false;

            // Else pop the corresponding
            // opening bracket from the stack
            else
                S.pop();
        }
    }

    // If no opening bracket remains
    // then return true
    if (S.size() == 0)
        return true;

    // If there are opening brackets
    // then return false
    else
        return false;
}

// Function to find number of
// strings having balanced brackets
void count(string ch, int index)
{

    // Reached end of character array
    if (index == ch.length()) {

        // Check if the character array
        // contains balanced string
        if (check(ch)) {

            // If it is a balanced string
            // then print its characters
            print(ch);
        }

        return;
    }

    if (ch[index] == '?') {

        // replace ? with (
        ch[index] = '(';

        count(ch, index + 1);

        // replace ? with )
        ch[index] = ')';

        count(ch, index + 1);

        // backtrack
        ch[index] = '?';
    }
    else {

        // If current character is a
        // valid bracket then continue
        // to the next character
        count(ch, index + 1);
    }
}

// Driver function
int main()
{

    string ch = "????";

    // Call the function
    count(ch, 0);
    return 0;
}

// This code is contributed by Potta Lokesh

Java

// Java implementation for the above approach

import java.io.*;
import java.util.*;

class Main {

    // Function to print the
    // characters of the string
    static void print(char ch[])
    {

        for (Character c : ch) {
            System.out.print(c);
        }
        System.out.println();
    }

    // Function to check if the
    // brackets are valid or not
    static boolean check(char ch[])
    {

        // Initialize a stack
        Stack<Character> S = new Stack<>();

        // If character is an open bracket
        // then return false
        if (ch[0] == ')') {

            return false;
        }

        // Iterate the character array
        for (int i = 0; i < ch.length; i++) {

            // If character is an open bracket
            // then push it in the stack
            if (ch[i] == '(') {

                S.add('(');
            }

            // If character is a close bracket
            else {

                // If stack is empty, there is no
                // corresponding opening bracket
                // so return false
                if (S.size() == 0)
                    return false;

                // Else pop the corresponding
                // opening bracket from the stack
                else
                    S.pop();
            }
        }

        // If no opening bracket remains
        // then return true
        if (S.size() == 0)
            return true;

        // If there are opening brackets
        // then return false
        else
            return false;
    }

    // Function to find number of
    // strings having balanced brackets
    static void count(char ch[], int index)
    {

        // Reached end of character array
        if (index == ch.length) {

            // Check if the character array
            // contains balanced string
            if (check(ch)) {

                // If it is a balanced string
                // then print its characters
                print(ch);
            }

            return;
        }

        if (ch[index] == '?') {

            // replace ? with (
            ch[index] = '(';

            count(ch, index + 1);

            // replace ? with )
            ch[index] = ')';

            count(ch, index + 1);

            // backtrack
            ch[index] = '?';
        }
        else {

            // If current character is a
            // valid bracket then continue
            // to the next character
            count(ch, index + 1);
        }
    }

    // Driver function
    public static void main(String[] args)
    {

        String m = "????";
        char ch[] = m.toCharArray();

        // Call the function
        count(ch, 0);
    }
}

Python3

# Python code for the above approach

# Function to print the
# characters of the string
def printf(ch):

  for c in ch:
    print(c, end="");
  print("");


# Function to check if the
# brackets are valid or not
def check(ch):

  # Initialize a stack
  S = [];

  # If character is an open bracket
  # then return false
  if (ch[0] == ')'):
    return False;


  # Iterate the character array
  for i in range(len(ch)):

    # If character is an open bracket
    # then push it in the stack
    if (ch[i] == '('):
      S.append('(');

    # If character is a close bracket
    else:

      # If stack is empty, there is no
      # corresponding opening bracket
      # so return false
      if (len(S) == 0):
        return False;

      # Else pop the corresponding
      # opening bracket from the stack
      else:
        S.pop();



  # If no opening bracket remains
  # then return true
  if (len(S) == 0):
    return True;

  # If there are opening brackets
  # then return false
  else:
    return False;


# Function to find number of
# strings having balanced brackets
def count(ch, index):

  # Reached end of character array
  if (index == len(ch)):

    # Check if the character array
    # contains balanced string
    if (check(ch)):

      # If it is a balanced string
      # then print its characters
      printf(ch);


    return;


  if (ch[index] == '?'):

    # replace ? with (
    ch[index] = '(';

    count(ch, index + 1);

    # replace ? with )
    ch[index] = ')';

    count(ch, index + 1);

    # backtrack
    ch[index] = '?';
  else:

    # If current character is a
    # valid bracket then continue
    # to the next character
    count(ch, index + 1);

# Driver function
ch = "????";

# Call the function
count(list(ch), 0);

# This code is contributed by Saurabh Jaiswal

C#

// C# implementation for the above approach
using System;
using System.Collections;

public class Gfg{

    // Function to print the
    // characters of the string
    static void print(char []ch)
    {

        foreach (char c in ch) {
            Console.Write(c);
        }
        Console.WriteLine();
    }

    // Function to check if the
    // brackets are valid or not
    static bool check(char []ch)
    {

        // Initialize a stack
        Stack S = new Stack();


        // If character is an open bracket
        // then return false
        if (ch[0] == ')') {

            return false;
        }

        // Iterate the character array
        for (int i = 0; i < ch.Length; i++) {

            // If character is an open bracket
            // then push it in the stack
            if (ch[i] == '(') {

                S.Push('(');
            }

            // If character is a close bracket
            else {

                // If stack is empty, there is no
                // corresponding opening bracket
                // so return false
                if (S.Count == 0)
                    return false;

                // Else pop the corresponding
                // opening bracket from the stack
                else
                    S.Pop();
            }
        }

        // If no opening bracket remains
        // then return true
        if (S.Count == 0)
            return true;

        // If there are opening brackets
        // then return false
        else
            return false;
    }

    // Function to find number of
    // strings having balanced brackets
    static void count(char []ch, int index)
    {

        // Reached end of character array
        if (index == ch.Length) {

            // Check if the character array
            // contains balanced string
            if (check(ch)) {

                // If it is a balanced string
                // then print its characters
                print(ch);
            }

            return;
        }

        if (ch[index] == '?') {

            // replace ? with (
            ch[index] = '(';

            count(ch, index + 1);

            // replace ? with )
            ch[index] = ')';

            count(ch, index + 1);

            // backtrack
            ch[index] = '?';
        }
        else {

            // If current character is a
            // valid bracket then continue
            // to the next character
            count(ch, index + 1);
        }
    }

    // Driver function
    public static void Main(string[] args)
    {

        string m = "????";
        char []ch = m.ToCharArray();

        // Call the function
        count(ch, 0);
    }
}

// This code is contributed by AnkThon

Javascript

<script>
// Javascript code for the above approach


// Function to print the
// characters of the string
function printf(ch) {
  for (c of ch) {
    document.write(c);
  }
  document.write("<br>");
}

// Function to check if the
// brackets are valid or not
function check(ch) {

  // Initialize a stack
  let S = [];

  // If character is an open bracket
  // then return false
  if (ch[0] == ')') {

    return false;
  }

  // Iterate the character array
  for (let i = 0; i < ch.length; i++) {

    // If character is an open bracket
    // then push it in the stack
    if (ch[i] == '(') {

      S.push('(');
    }

    // If character is a close bracket
    else {

      // If stack is empty, there is no
      // corresponding opening bracket
      // so return false
      if (S.length == 0)
        return false;

      // Else pop the corresponding
      // opening bracket from the stack
      else
        S.pop();
    }
  }

  // If no opening bracket remains
  // then return true
  if (S.length == 0)
    return true;

  // If there are opening brackets
  // then return false
  else
    return false;
}

// Function to find number of
// strings having balanced brackets
function count(ch, index) {

  // Reached end of character array
  if (index == ch.length) {

    // Check if the character array
    // contains balanced string
    if (check(ch)) {

      // If it is a balanced string
      // then print its characters
      printf(ch);
    }

    return;
  }

  if (ch[index] == '?') {

    // replace ? with (
    ch[index] = '(';

    count(ch, index + 1);

    // replace ? with )
    ch[index] = ')';

    count(ch, index + 1);

    // backtrack
    ch[index] = '?';
  }
  else {

    // If current character is a
    // valid bracket then continue
    // to the next character
    count(ch, index + 1);
  }
}

// Driver function
let ch = "????";

// Call the function
count(ch.split(""), 0);

// This code is contributed by Saurabh Jaiswal
</script>

Output

(())
()()

Time Complexity: O(N*2^N)
Auxiliary Space: O(N)