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Check if a given string can be converted to a Balanced Bracket Sequence
  • Difficulty Level : Medium
  • Last Updated : 13 Jan, 2021

Given a string S of size N consisting of ‘(‘, ‘)’, and ‘$’, the task is to check whether the given string can be converted into a balanced bracket sequence by replacing every occurrence of $ with either ) or (.

A balanced bracket sequence is sequence where every opening bracket “(“ has a corresponding closing bracket “)”.

Examples:

Input: S = “()($”
Output: Yes
Explanation: Convert the string into balanced bracket sequence: ()().

Input: S = “$()$(“
Output: No
Explanation: Possible replacements are “(((((“, “(())(“, “)(()(“, “)()((“, none of which are balanced. Hence, a balanced bracket sequence can not be obtained.



Approach: The above problem can be solved by using a Stack. The idea is to check if all ) can be balanced with ( or $  and vice versa. Follow the steps below to solve this problem:

  • Store the frequency of “(“, “)” and “$” in a variable as countOpen, countClosed and countSymbol respectively.
  • Initialize a variable ans as 1 to store the required result and a stack stack_1 to check if all “)” can be balanced with “(“ or $.
  • Traverse the string S using the variable i and do the following:
  • Reverse the string S, and follow the same procedure to check if all “(“ can be balanced with “)” or “$”.
  • If the value of countSymbol is less than the absolute difference of countOpen and countClosed then set ans to 0. Else balance the extra parenthesis with the symbols. After balancing if countSymbol is odd, set ans as 0.
  • After the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if the string
// can be balanced by replacing the
// '$' with opening or closing brackets
bool canBeBalanced(string sequence)
{
    // If string can never be balanced
    if (sequence.size() % 2)
        return false;
 
    // Declare 2 stacks to check if all
    // ) can be balanced with ( or $
    // and vice-versa
    stack<char> stack_, stack2_;
 
    // Store the count the occurence
    // of (, ) and $
    int countOpen = 0, countClosed = 0;
    int countSymbol = 0;
 
    // Traverse the string
    for (int i = 0;
         i < sequence.size(); i++) {
 
        if (sequence[i] == ')') {
 
            // Increment closed bracket
            // count by 1
            countClosed++;
 
            // If there are no opening
            // bracket to match it
            // then return false
            if (stack_.empty()) {
                return false;
            }
 
            // Otherwise, pop the character
            // from the stack
            else {
                stack_.pop();
            }
        }
 
        else {
 
            // If current character is
            // an opening bracket or $,
            // push it to the stack
            if (sequence[i] == '$') {
 
                // Increment symbol
                // count by 1
                countSymbol++;
            }
            else {
 
                // Increment open
                // bracket count by 1
                countOpen++;
            }
            stack_.push(sequence[i]);
        }
    }
 
    // Traverse the string from end
    // and repeat the same process
    for (int i = sequence.size() - 1;
         i >= 0; i--) {
 
        if (sequence[i] == '(') {
 
            // If there are no closing
            // brackets to match it
            if (stack2_.empty()) {
                return false;
            }
 
            // Otherwise, pop character
            // from stack
            else {
                stack2_.pop();
            }
        }
        else {
            stack2_.push(sequence[i]);
        }
    }
 
    // Store the extra ( or ) which
    // are not balanced yet
    int extra = abs(countClosed - countOpen);
 
    // Check if $ is available to
    // balance the extra brackets
    if (countSymbol < extra) {
        return false;
    }
 
    else {
 
        // Count ramaining $ after
        // balancing extra ( and )
        countSymbol -= extra;
 
        // Check if each pair of $
        // is convertable in ()
        if (countSymbol % 2 == 0) {
            return true;
        }
    }
    return false;
}
 
// Driver Code
int main()
{
    string S = "()($";
 
    // Function Call
    if (canBeBalanced(S)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to check if the String
// can be balanced by replacing the
// '$' with opening or closing brackets
static boolean canBeBalanced(String sequence)
{
   
    // If String can never be balanced
    if (sequence.length() % 2 == 1)
        return false;
 
    // Declare 2 stacks to check if all
    // ) can be balanced with ( or $
    // and vice-versa
    Stack<Character> stack_ = new Stack<Character>();
    Stack<Character> stack2_ = new Stack<Character>();
 
    // Store the count the occurence
    // of (, ) and $
    int countOpen = 0, countClosed = 0;
    int countSymbol = 0;
 
    // Traverse the String
    for (int i = 0;
         i < sequence.length(); i++)
    {
 
        if (sequence.charAt(i) == ')')
        {
 
            // Increment closed bracket
            // count by 1
            countClosed++;
 
            // If there are no opening
            // bracket to match it
            // then return false
            if (stack_.isEmpty())
            {
                return false;
            }
 
            // Otherwise, pop the character
            // from the stack
            else
            {
                stack_.pop();
            }
        }
 
        else
        {
 
            // If current character is
            // an opening bracket or $,
            // push it to the stack
            if (sequence.charAt(i) == '$')
            {
 
                // Increment symbol
                // count by 1
                countSymbol++;
            }
            else
            {
 
                // Increment open
                // bracket count by 1
                countOpen++;
            }
            stack_.add(sequence.charAt(i));
        }
    }
 
    // Traverse the String from end
    // and repeat the same process
    for (int i = sequence.length() - 1;
         i >= 0; i--)
    {
 
        if (sequence.charAt(i) == '(')
        {
 
            // If there are no closing
            // brackets to match it
            if (stack2_.isEmpty())
            {
                return false;
            }
 
            // Otherwise, pop character
            // from stack
            else
            {
                stack2_.pop();
            }
        }
        else
        {
            stack2_.add(sequence.charAt(i));
        }
    }
 
    // Store the extra ( or ) which
    // are not balanced yet
    int extra = Math.abs(countClosed - countOpen);
 
    // Check if $ is available to
    // balance the extra brackets
    if (countSymbol < extra)
    {
        return false;
    }
 
    else
    {
 
        // Count ramaining $ after
        // balancing extra ( and )
        countSymbol -= extra;
 
        // Check if each pair of $
        // is convertable in ()
        if (countSymbol % 2 == 0)
        {
            return true;
        }
    }
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    String S = "()($";
 
    // Function Call
    if (canBeBalanced(S))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by 29AjayKumar

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Python3

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# Python3 program for the above approach
 
# Function to check if the
# can be balanced by replacing the
# '$' with opening or closing brackets
def canBeBalanced(sequence):
     
    # If string can never be balanced
    if (len(sequence) % 2):
        return False
 
    # Declare 2 stacks to check if all
    # ) can be balanced with ( or $
    # and vice-versa
    stack_, stack2_ = [], []
 
    # Store the count the occurence
    # of (, ) and $
    countOpen ,countClosed = 0, 0
    countSymbol = 0
 
    # Traverse the string
    for i in range(len(sequence)):
        if (sequence[i] == ')'):
 
            # Increment closed bracket
            # count by 1
            countClosed += 1
 
            # If there are no opening
            # bracket to match it
            # then return False
            if (len(stack_) == 0):
                return False
 
            # Otherwise, pop the character
            # from the stack
            else:
                del stack_[-1]
        else:
 
            # If current character is
            # an opening bracket or $,
            # push it to the stack
            if (sequence[i] == '$'):
 
                # Increment symbol
                # count by 1
                countSymbol += 1
            else:
 
                # Increment open
                # bracket count by 1
                countOpen += 1
            stack_.append(sequence[i])
 
    # Traverse the string from end
    # and repeat the same process
    for i in range(len(sequence)-1, -1, -1):
        if (sequence[i] == '('):
 
            # If there are no closing
            # brackets to match it
            if (len(stack2_) == 0):
                return False
 
            # Otherwise, pop character
            # from stack
            else:
                del stack2_[-1]
        else :
            stack2_.append(sequence[i])
 
    # Store the extra ( or ) which
    # are not balanced yet
    extra = abs(countClosed - countOpen)
 
    # Check if $ is available to
    # balance the extra brackets
    if (countSymbol < extra):
        return False
    else :
 
        # Count ramaining $ after
        # balancing extra ( and )
        countSymbol -= extra
 
        # Check if each pair of $
        # is convertable in ()
        if (countSymbol % 2 == 0) :
            return True
    return False
 
# Driver Code
if __name__ == '__main__':
    S = "()($"
 
    # Function Call
    if (canBeBalanced(S)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by mohit kumar 29

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C#

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// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to check if the String
  // can be balanced by replacing the
  // '$' with opening or closing brackets
  static bool canBeBalanced(String sequence)
  {
 
    // If String can never be balanced
    if (sequence.Length % 2 == 1)
      return false;
 
    // Declare 2 stacks to check if all
    // ) can be balanced with ( or $
    // and vice-versa
    Stack<char> stack_ = new Stack<char>();
    Stack<char> stack2_ = new Stack<char>();
 
    // Store the count the occurence
    // of (, ) and $
    int countOpen = 0, countClosed = 0;
    int countSymbol = 0;
 
    // Traverse the String
    for (int i = 0;
         i < sequence.Length; i++)
    {
      if (sequence[i] == ')')
      {
 
        // Increment closed bracket
        // count by 1
        countClosed++;
 
        // If there are no opening
        // bracket to match it
        // then return false
        if (stack_.Count==0)
        {
          return false;
        }
 
        // Otherwise, pop the character
        // from the stack
        else
        {
          stack_.Pop();
        }
      }
 
      else
      {
 
        // If current character is
        // an opening bracket or $,
        // push it to the stack
        if (sequence[i] == '$')
        {
 
          // Increment symbol
          // count by 1
          countSymbol++;
        }
        else
        {
 
          // Increment open
          // bracket count by 1
          countOpen++;
        }
        stack_.Push(sequence[i]);
      }
    }
 
    // Traverse the String from end
    // and repeat the same process
    for (int i = sequence.Length - 1;
         i >= 0; i--)
    {
 
      if (sequence[i] == '(')
      {
 
        // If there are no closing
        // brackets to match it
        if (stack2_.Count == 0)
        {
          return false;
        }
 
        // Otherwise, pop character
        // from stack
        else
        {
          stack2_.Pop();
        }
      }
      else
      {
        stack2_.Push(sequence[i]);
      }
    }
 
    // Store the extra ( or ) which
    // are not balanced yet
    int extra = Math.Abs(countClosed - countOpen);
 
    // Check if $ is available to
    // balance the extra brackets
    if (countSymbol < extra)
    {
      return false;
    }
 
    else
    {
 
      // Count ramaining $ after
      // balancing extra ( and )
      countSymbol -= extra;
 
      // Check if each pair of $
      // is convertable in ()
      if (countSymbol % 2 == 0)
      {
        return true;
      }
    }
    return false;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    String S = "()($";
 
    // Function Call
    if (canBeBalanced(S))
    {
      Console.Write("Yes");
    }
    else
    {
      Console.Write("No");
    }
  }
}
 
// This code is contributed by 29AjayKumar

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Output: 

Yes

 

Time Complexity: O(N)
Auxiliary Space: O(N)

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