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

• Difficulty Level : Easy
• Last Updated : 13 Jul, 2022

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

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 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 balanced``    ``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 balanced``        ``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 balanced``    ``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 balanced``        ``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`

## Javascript

 ``

Output:

`Yes`

Time Complexity : O(n) ,where n is size of given string

Space Complexity : O(n)

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