# 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 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 ` `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 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 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"``) `

## C#

 `// 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 `

Output:

```Yes
```

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.