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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// 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 

Java

// 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 

Python3

# 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") 

Output:

Yes

My Personal Notes

Suggest a Topic

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

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