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 which 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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

Yes

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