Check if the bracket sequence can be balanced with at most one change in the position of a bracket

Given an unbalanced bracket sequence as a string str, the task is to find whether the given string can be balanced by moving at most one bracket from its original place in the sequence to any other position.

Examples:

Input: str = “)(()”
Output: Yes
As by moving s[0] to the end will make it valid.
“(())”



Input: str = “()))(()”
Output: No

Approach: Consider X as a valid bracket then definitely (X) is also valid. If X is not valid and can be balanced with just one change of position in some bracket then it must be of the type X = “)(“ where ‘)’ has been placed before ‘(‘.
Now, X can be replaced with (X) as it will not affect the balanced nature of X. The new string becomes X = “()()” which is balanced.
Hence, if (X) is balanced then we can say that X can be balanced with at most one change in the position of some bracket.

Below is the implementation of the above approach:

C++

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// CPP implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if the sequence
// can be balanced by changing the
// position of at most one bracket
bool canBeBalanced(string s, int n)
{
    // Odd length string can
    // never be balnced
    if (n % 2 == 1)
        return false;
  
    // Add '(' in the beginning and ')'
    // in the end of the string
    string k = "(";
    k += s + ")";
  
    vector<string> d;
    int cnt = 0;
  
    for (int i = 0; i < k.length(); i++)
    {
        // If its an opening bracket then
        // append it to the temp string
        if (k[i] == '(')
            d.push_back("(");
  
        // If its a closing bracket
        else
        {
            // There was an opening bracket
            // to match it with
            if (d.size() != 0)
                d.pop_back();
  
            // No opening bracket to
            // match it with
            else
                return false;
        }
    }
  
    // Sequence is balanced
    if (d.empty())
        return true;
    return false;
}
  
// Driver Code
int main(int argc, char const *argv[])
{
    string s = ")(()";
    int n = s.length();
  
    (canBeBalanced(s, n)) ? cout << "Yes" 
                  << endl : cout << "No" << endl;
    return 0;
}
  
// This code is contributed by
// sanjeev2552

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Java














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// Java implementation of the approach 


import java.util.Vector; 


  


class GFG  


{ 


  


    // Function that returns true if the sequence 


    // can be balanced by changing the 


    // position of at most one bracket 


    static boolean canBeBalanced(String s, int n) 


    { 


  


        // Odd length string can 


        // never be balnced 


        if (n % 2 == 1) 


            return false; 


  


        // Add '(' in the beginning and ')' 


        // in the end of the string 


        String k = "("; 


        k += s + ")"; 


        Vector<String> d = new Vector<>(); 


  


        for (int i = 0; i < k.length(); i++) 


        { 


  


            // If its an opening bracket then 


            // append it to the temp string 


            if (k.charAt(i) == '(') 


                d.add("("); 


  


            // If its a closing bracket 


            else 


            { 


  


                // There was an opening bracket 


                // to match it with 


                if (d.size() != 0) 


                    d.remove(d.size() - 1); 


  


                // No opening bracket to 


                // match it with 


                else


                    return false; 


            } 


        } 


  


        // Sequence is balanced 


        if (d.isEmpty()) 


            return true; 


        return false; 


    } 


  


    // Driver Code 


    public static void main(String[] args)  


    { 


        String s = ")(()"; 


        int n = s.length(); 


  


        if (canBeBalanced(s, n)) 


            System.out.println("Yes"); 


        else


            System.out.println("No"); 


    } 


} 


  


// This code is contributed by 


// sanjeev2552 



















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Python3














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# Python3 implementation of the approach 


  


# Function that returns true if the sequence  


# can be balanced by changing the  


# position of at most one bracket 


def canBeBalanced(s, n): 


  


    # Odd length string can  


    # never be balnced 


    if n % 2 == 1: 


        return False


  


    # Add '(' in the beginning and ')'  


    # in the end of the string 


    k = "("


    k = k + s+")"


    d = [] 


    count = 0


    for i in range(len(k)): 


  


        # If its an opening bracket then  


        # append it to the temp string 


        if k[i] == "(": 


            d.append("(") 


  


        # If its a closing bracket 


        else: 


  


            # There was an opening bracket  


            # to match it with 


            if len(d)!= 0: 


                d.pop() 


  


            # No opening bracket to  


            # match it with 


            else: 


                return False


      


    # Sequence is balanced 


    if len(d) == 0: 


        return True


    return False


  


# Driver code 


S = ")(()"


n = len(S) 


if(canBeBalanced(S, n)): 


    print("Yes") 


else: 


    print("No") 



















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C#














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// C# implementation of the approach 


using System; 


using System.Collections.Generic; 


  


class GFG  


{ 


  


    // Function that returns true if the sequence 


    // can be balanced by changing the 


    // position of at most one bracket 


    static bool canBeBalanced(string s, int n) 


    { 


  


        // Odd length string can 


        // never be balnced 


        if (n % 2 == 1) 


            return false; 


  


        // Add '(' in the beginning and ')' 


        // in the end of the string 


        string k = "("; 


        k += s + ")"; 


        List<string> d = new List<string>(); 


  


        for (int i = 0; i < k.Length; i++) 


        { 


  


            // If its an opening bracket then 


            // append it to the temp string 


            if (k[i] == '(') 


                d.Add("("); 


  


            // If its a closing bracket 


            else


            { 


  


                // There was an opening bracket 


                // to match it with 


                if (d.Count != 0) 


                    d.RemoveAt(d.Count - 1); 


  


                // No opening bracket to 


                // match it with 


                else


                    return false; 


            } 


        } 


  


        // Sequence is balanced 


        if (d.Count == 0) 


            return true; 


        return false; 


    } 


  


    // Driver Code 


    public static void Main()  


    { 


        string s = ")(()"; 


        int n = s.Length; 


  


        if (canBeBalanced(s, n)) 


            Console.Write("Yes"); 


        else


            Console.Write("No"); 


    } 


} 


  


// This code is contributed by 


// mohit kumar 29 



















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Output:


Yes






          
          
          
          

            

    

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    Article Tags : 
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Practice Tags : 
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