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Check if the bracket sequence can be balanced with at most one change in the position of a bracket | Set 2
  • Last Updated : 04 May, 2021

Given a 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: The problem can be solved using a stack as discussed in this post. In this article, an approach that doesn’t use extra space will be discussed. 
If the frequency of ‘(‘ is less than frequency of ‘)’. If the above difference is greater than 1 then the sequence cannot be balanced else it can be balanced if the overall difference is zero.
Below is implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if
// the string can be balanced
bool canBeBalanced(string s, int n)
{
 
    // Count to check the difference between
    // the frequencies of '(' and ')' and
    // count_1 is to find the minimum value
    // of freq('(') - freq(')')
    int count = 0, count_1 = 0;
 
    // Traverse the given string
    for (int i = 0; i < n; i++) {
 
        // Increase the count
        if (s[i] == '(')
            count++;
 
        // Decrease the count
        else
            count--;
 
        // Find the minimum value
        // of freq('(') - freq(')')
        count_1 = min(count_1, count);
    }
 
    // If the minimum difference is greater
    // than or equal to -1 and the overall
    // difference is zero
    if (count_1 >= -1 && count == 0)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    string s = "())()(";
    int n = s.length();
 
    if (canBeBalanced(s, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java program to toggle K-th bit of a number N
class GFG
{
 
// Function that returns true if
// the string can be balanced
static boolean canBeBalanced(String s, int n)
{
 
    // Count to check the difference between
    // the frequencies of '(' and ')' and
    // count_1 is to find the minimum value
    // of freq('(') - freq(')')
    int count = 0, count_1 = 0;
 
    // Traverse the given string
    for (int i = 0; i < n; i++)
    {
 
        // Increase the count
        if (s.charAt(i) == '(')
            count++;
 
        // Decrease the count
        else
            count--;
 
        // Find the minimum value
        // of freq('(') - freq(')')
        count_1 = Math.min(count_1, count);
    }
 
    // If the minimum difference is greater
    // than or equal to -1 and the overall
    // difference is zero
    if (count_1 >= -1 && count == 0)
        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 PrinciRaj1992

Python3




# Python3 implementation of the approach
 
# Function that returns true if
# the can be balanced
def canBeBalanced(s, n):
 
    # Count to check the difference between
    # the frequencies of '(' and ')' and
    # count_1 is to find the minimum value
    # of freq('(') - freq(')')
    count = 0
    count_1 = 0
 
    # Traverse the given string
    for i in range(n):
 
        # Increase the count
        if (s[i] == '('):
            count += 1
 
        # Decrease the count
        else:
            count -= 1
 
        # Find the minimum value
        # of freq('(') - freq(')')
        count_1 = min(count_1, count)
 
    # If the minimum difference is greater
    # than or equal to -1 and the overall
    # difference is zero
    if (count_1 >= -1 and count == 0):
        return True
 
    return False
 
# Driver code
s = "())()("
n = len(s)
 
if (canBeBalanced(s, n)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Mohit Kumar

C#




// C# program to toggle K-th bit of a number N
using System;
 
class GFG
{
 
// Function that returns true if
// the string can be balanced
static Boolean canBeBalanced(String s, int n)
{
 
    // Count to check the difference between
    // the frequencies of '(' and ')' and
    // count_1 is to find the minimum value
    // of freq('(') - freq(')')
    int count = 0, count_1 = 0;
 
    // Traverse the given string
    for (int i = 0; i < n; i++)
    {
 
        // Increase the count
        if (s[i] == '(')
            count++;
 
        // Decrease the count
        else
            count--;
 
        // Find the minimum value
        // of freq('(') - freq(')')
        count_1 = Math.Min(count_1, count);
    }
 
    // If the minimum difference is greater
    // than or equal to -1 and the overall
    // difference is zero
    if (count_1 >= -1 && count == 0)
        return true;
 
    return false;
}
 
// Driver code
public static void Main(String []args)
{
    String s = "())()(";
    int n = s.Length;
 
    if (canBeBalanced(s, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript program to find
// next greater number than N
 
 
// Function that returns true if
// the string can be balanced
function canBeBalanced( s, n)
{
 
    // Count to check the difference between
    // the frequencies of '(' and ')' and
    // count_1 is to find the minimum value
    // of freq('(') - freq(')')
    var count = 0, count_1 = 0;
 
    // Traverse the given string
    for (var i=0; i < n; i++) {
 
        // Increase the count
        if (s[i] == '(')
            count++;
 
        // Decrease the count
        else
            count--;
 
        // Find the minimum value
        // of freq('(') - freq(')')
        count_1 = Math.min(count_1, count);
    }
 
    // If the minimum difference is greater
    // than or equal to -1 and the overall
    // difference is zero
    if (count_1 >= -1 && count == 0)
        return true;
 
    return false;
}
 
var s = "())()(";
    var n = s.length;
 
    if (canBeBalanced(s, n))
        document.write("Yes");
    else
        document.write("No");
 
// This code is contributed bby SoumikMondal
 
</script>
Output: 
Yes

 

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