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Modify a numeric string to a balanced parentheses by replacements

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  • Last Updated : 08 Sep, 2021
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Given a numeric string S made up of characters ‘1’, ‘2’ and ‘3’ only, the task is to replace characters with either an open bracket ( ‘(‘ ) or a closed bracket ( ‘)’ ) such that the newly formed string becomes a balanced bracket sequence.

Note: All occurrences of a character must be replaced by the same parentheses.

Examples:

Input: S = “1123”
Output: Yes, (())
Explanation: Replacing occurrences of character ‘1’ with ‘(‘, ‘2’ with ‘)’ and ‘3’ with ‘)’. Therefore, the obtained bracket sequence is “(())”, which is balanced.

Input: S = “1121”
Output: No

Approach: The given problem can be solved based on the following observations:

  • For a balanced bracket sequence, it is necessary for the first and last characters to be open and closed brackets respectively. Therefore, the first and the last characters should be different.
  • If the first and the last characters of a string are the same, then it is impossible to obtain a balanced bracket sequence.
  • If the first and last characters of a string are different, then they are replaced by open and closed brackets respectively. The third character is replaced either by open or closed brackets.
  • Check for both ways of replacements one by one for the remaining third character.
  • If both replacements of the third remaining character can’t make a balanced bracket sequence, then it is impossible to make a balanced bracket sequence.

Follow the steps below to solve the given problem:

  • Check if the first and last characters of the string S are equal or not. If found to be true, then print “No” and return.
  • Initialize two variables, say cntforOpen and cntforClose, to store the count of open and closed brackets.
  • Iterate over the characters of the string and perform the following operations:
    • If the current character is the same as the first character of the string, increment cntforOpen.
    • If the current character is the same as the last character of the string, decrement cntforOpen.
    • For the remaining third character, increment cntforOpen, i.e. replacing that character with ‘(‘.
    • If at any instant, cntforOpen is found to be negative, then a balanced bracket sequence cannot be obtained.
  • Similarly, check using cntforClose variable, i.e. replacing the third character with ‘)’.
  • If none of the above two methods generates a balanced bracket sequence, then print “No”. Otherwise, print “Yes”.

Below is the implementation of the above approach:  

C++




// C++ program for the above approach
#include <iostream>
using namespace std;
 
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
void balBracketSequence(string str)
{
    int n = str.size();
 
    // Check if the first and
    // last characters are equal
    if (str[0]
        == str[n - 1])
 
    {
        cout << "No" << endl;
    }
    else {
 
        // Initialize two variables to store
        // the count of open and closed brackets
        int cntForOpen = 0, cntForClose = 0;
        int check = 1;
        for (int i = 0; i < n; i++) {
 
            // If the current character is
            // same as the first character
            if (str[i] == str[0])
                cntForOpen++;
 
            // If the current character is
            // same as the last character
            else if (str[i] == str[n - 1])
                cntForOpen--;
            else
                cntForOpen++;
 
            // If count of open brackets
            // becomes less than 0
            if (cntForOpen < 0) {
                check = 0;
                break;
            }
        }
        if (check && cntForOpen == 0) {
            cout << "Yes, ";
 
            // Print the new string
            for (int i = 0; i < n; i++) {
                if (str[i] == str[n - 1])
                    cout << ')';
                else
                    cout << '(';
            }
            return;
        }
        else {
            for (int i = 0; i < n; i++) {
 
                // If the current character is
                // same as the first character
                if (str[i] == str[0])
                    cntForClose++;
                else
                    cntForClose--;
 
                // If bracket sequence
                // is not balanced
                if (cntForClose
                    < 0) {
                    check = 0;
                    break;
                }
            }
 
            // Check for unbalanced
            // bracket sequence
            if (check
                && cntForClose
                       == 0) {
                cout << "Yes, ";
 
                // Print the sequence
                for (int i = 0; i < n;
                     i++) {
                    if (str[i] == str[0])
                        cout << '(';
                    else
                        cout << ')';
                }
                return;
            }
        }
        cout << "No";
    }
}
 
// Driver Code
int main()
{
    // Given Input
    string str = "123122";
 
    // Function Call
    balBracketSequence(str);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
static void balBracketSequence(String str)
{
    int n = str.length();
 
    // Check if the first and
    // last characters are equal
    if (str.charAt(0)
        == str.charAt(n - 1))
 
    {
        System.out.println("No");
    }
    else {
 
        // Initialize two variables to store
        // the count of open and closed brackets
        int cntForOpen = 0, cntForClose = 0;
        int check = 1;
        for (int i = 0; i < n; i++) {
 
            // If the current character is
            // same as the first character
            if (str.charAt(i) == str.charAt(0))
                cntForOpen++;
 
            // If the current character is
            // same as the last character
            else if (str.charAt(i) == str.charAt(n - 1))
                cntForOpen -= 1;
            else
                cntForOpen += 1;
 
            // If count of open brackets
            // becomes less than 0
            if (cntForOpen < 0) {
                check = 0;
                break;
            }
        }
        if (check != 0 && cntForOpen == 0) {
            System.out.print("Yes, ");
 
            // Print the new string
            for (int i = 0; i < n; i++) {
                if (str.charAt(i) == str.charAt(n - 1))
                    System.out.print(')');
                else
                   System.out.print('(');
            }
            return;
        }
        else {
            for (int i = 0; i < n; i++) {
 
                // If the current character is
                // same as the first character
                if (str.charAt(i) == str.charAt(0))
                    cntForClose++;
                else
                    cntForClose--;
 
                // If bracket sequence
                // is not balanced
                if (cntForClose
                    < 0) {
                    check = 0;
                    break;
                }
            }
 
            // Check for unbalanced
            // bracket sequence
            if (check != 0
                && cntForClose
                       == 0) {
                System.out.print("Yes, ");
 
                // Print the sequence
                for (int i = 0; i < n;
                     i++) {
                    if (str.charAt(i) == str.charAt(0))
                       System.out.print('(');
                    else
                        System.out.print(')');
                }
                return;
            }
        }
        System.out.print("No");
    }
}
 
// Driver Code
public static void main(String args[])
{
    // Given Input
    String str = "123122";
 
    // Function Call
    balBracketSequence(str);
}
}
 
// This code is contributed by ipg2016107.

Python3




# Python program for the above approach;
# Function to check if the given
# string can be converted to a
# balanced bracket sequence or not
def balBracketSequence(str):
    n = len(str)
 
    # Check if the first and
    # last characters are equal
    if (str[0] == str[n - 1]):
        print("No", end="")
    else:
 
        # Initialize two variables to store
        # the count of open and closed brackets
        cntForOpen = 0
        cntForClose = 0
        check = 1
 
        for i in range(n):
 
            # If the current character is
            # same as the first character
            if (str[i] == str[0]):
                cntForOpen += 1
 
            # If the current character is
            # same as the last character
            elif str[i] == str[n - 1] :
                cntForOpen -= 1
            else:
                cntForOpen += 1
 
            # If count of open brackets
            # becomes less than 0
            if (cntForOpen < 0):
                check = 0
                break
             
         
        if (check and cntForOpen == 0):
            print("Yes, ", end="")
 
            # Print the new string
            for i in range(n):
                if (str[i] == str[n - 1]):
                    print(')', end="")
                else:
                    print('(', end="")
             
            return
         
        else:
            for i in range(n):
 
                # If the current character is
                # same as the first character
                if (str[i] == str[0]):
                    cntForClose += 1
                else:
                    cntForClose -= 1
 
                # If bracket sequence
                # is not balanced
                if (cntForClose < 0):
                    check = 0
                    break
                 
             
 
            # Check for unbalanced
            # bracket sequence
            if (check and cntForClose == 0):
                print("Yes, ", end="")
 
                # Print the sequence
                for i in range(n):
                    if (str[i] == str[0]):
                        print('(', end="") 
                    else:
                        print(')', end="")
                 
                return
             
         
        print("NO", end="")
     
 
 
# Driver Code
 
# Given Input
str = "123122"
 
# Function Call
balBracketSequence(str)
 
# This code is contributed by gfgking

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
static void balBracketSequence(string str)
{
    int n = str.Length;
     
    // Check if the first and
    // last characters are equal
    if (str[0] == str[n - 1])
    {
        Console.Write("No");
    }
    else
    {
         
        // Initialize two variables to store
        // the count of open and closed brackets
        int cntForOpen = 0, cntForClose = 0;
        int check = 1;
         
        for(int i = 0; i < n; i++)
        {
             
            // If the current character is
            // same as the first character
            if (str[i] == str[0])
                cntForOpen++;
 
            // If the current character is
            // same as the last character
            else if (str[i] == str[n - 1])
                cntForOpen--;
            else
                cntForOpen++;
 
            // If count of open brackets
            // becomes less than 0
            if (cntForOpen < 0)
            {
                check = 0;
                break;
            }
        }
        if (check != 0 && cntForOpen == 0)
        {
            Console.Write("Yes, ");
 
            // Print the new string
            for(int i = 0; i < n; i++)
            {
                if (str[i] == str[n - 1])
                    Console.Write(')');
                else
                    Console.Write('(');
            }
            return;
        }
        else
        {
            for(int i = 0; i < n; i++)
            {
                 
                // If the current character is
                // same as the first character
                if (str[i] == str[0])
                    cntForClose++;
                else
                    cntForClose--;
 
                // If bracket sequence
                // is not balanced
                if (cntForClose < 0)
                {
                    check = 0;
                    break;
                }
            }
 
            // Check for unbalanced
            // bracket sequence
            if (check != 0 && cntForClose == 0)
            {
                Console.Write("Yes, ");
 
                // Print the sequence
                for(int i = 0; i < n; i++)
                {
                    if (str[i] == str[0])
                        Console.Write('(');
                    else
                        Console.Write(')');
                }
                return;
            }
        }
        Console.Write("No");
    }
}
 
// Driver Code
public static void Main()
{
     
    // Given Input
    string str = "123122";
 
    // Function Call
    balBracketSequence(str);
}
}
 
// This code is contributed by sanjoy_62

Javascript




<script>
 
        // JavaScript program for the above approach;
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
function balBracketSequence(str)
{
    let n = str.length;
 
    // Check if the first and
    // last characters are equal
    if (str[0]
        == str[n - 1])
 
    {
        document.write( "No");
    }
    else {
 
        // Initialize two variables to store
        // the count of open and closed brackets
        let cntForOpen = 0, cntForClose = 0;
        let check = 1;
        for (let i = 0; i < n; i++) {
 
            // If the current character is
            // same as the first character
            if (str[i] == str[0])
                cntForOpen++;
 
            // If the current character is
            // same as the last character
            else if (str[i] == str[n - 1])
                cntForOpen--;
            else
                cntForOpen++;
 
            // If count of open brackets
            // becomes less than 0
            if (cntForOpen < 0) {
                check = 0;
                break;
            }
        }
        if (check && cntForOpen == 0) {
            document.write("Yes, ");
 
            // Print the new string
            for (let i = 0; i < n; i++) {
                if (str[i] == str[n - 1])
                    document.write(')');
                else
                    document.write('(');
            }
            return;
        }
        else {
            for (let i = 0; i < n; i++) {
 
                // If the current character is
                // same as the first character
                if (str[i] == str[0])
                    cntForClose++;
                else
                    cntForClose--;
 
                // If bracket sequence
                // is not balanced
                if (cntForClose
                    < 0) {
                    check = 0;
                    break;
                }
            }
 
            // Check for unbalanced
            // bracket sequence
            if (check
                && cntForClose
                       == 0) {
                document.write("Yes, ");
 
                // Print the sequence
                for (let i = 0; i < n;
                     i++) {
                    if (str[i] == str[0])
                        document.write('(');
                    else
                        document.write(')');
                }
                return;
            }
        }
       document.write("NO") ;
    }
}
 
// Driver Code
 
    // Given Input
    let str = "123122";
 
    // Function Call
    balBracketSequence(str);
    
   // This code is contributed by Potta Lokesh
    </script>

Output: 

Yes, ()(())

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 


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