Minimum number of bracket reversals needed to make an expression balanced

Given an expression with only ‘}’ and ‘{‘. The expression may not be balanced. Find minimum number of bracket reversals to make the expression balanced.
Examples: 

Input:  exp = "}{"
Output: 2
We need to change '}' to '{' and '{' to
'}' so that the expression becomes balanced, 
the balanced expression is '{}'
Input:  exp = "{{{"
Output: Can't be made balanced using reversals
Input:  exp = "{{{{"
Output: 2 
Input:  exp = "{{{{}}"
Output: 1 
Input:  exp = "}{{}}{{{"
Output: 3

One simple observation is, the string can be balanced only if total number of brackets is even (there must be equal no of ‘{‘ and ‘}’)
A Naive Solution is to consider every bracket and recursively count number of reversals by taking two cases (i) keeping the bracket as it is (ii) reversing the bracket. If we get a balanced expression, we update result if number of steps followed for reaching here is smaller than the minimum so far. 

Steps to implement

• Check for the length of the given expression, if it is odd then return -1
• If it is even then call a recursive function
• That recursive function once leaves a bracket as it is and once reverses a bracket
• After each recursive call, we will check whether the modified string is balanced or not
  • If the modified string is balanced and the number of reversals to reach this string is minimum from all the answers found till now then keep this answer stored somewhere
• In the function for checking whether the string is balanced or not, we will initially keep count=0.
  • After that traverse the string and if we got ‘{‘ then increment the count else decrement the count
  • If at any point count becomes negative then this means there are more Closing brackets than opening ones. Hence string is not balanced
  • At last, if the count is not equal to 0 then that means there are more opening brackets than closing ones. Hence string is not balanced
  • If we never found that there are more Closing brackets than opening ones or there are more opening brackets than closing ones then the string is balanced

Code-

Output-

2

Time complexity: O(2ⁿ*n), because O(2ⁿ) in recursive function and O(n) in checking the expression generated after every recursive call is balanced or not

Space Complexity: O(n), because of the Auxillary space of recursion

An Efficient Solution can solve this problem in O(n) time. The idea is to first remove all balanced part of expression. For example, convert “}{{}}{{{” to “}{{{” by removing highlighted part. If we take a closer look, we can notice that, after removing balanced part, we always end up with an expression of the form }}…}{{…{, an expression that contains 0 or more number of closing brackets followed by 0 or more numbers of opening brackets. 
How many minimum reversals are required for an expression of the form “}}..}{{..{” ?. Let m be the total number of closing brackets and n be the number of opening brackets. We need &#x2308m/2&#x2309 + &#x2308n/2&#x2309 reversals. For example }}}}{{ requires 2+1 reversals.
Below is implementation of above idea: 

2

Time Complexity: O(n) 
Auxiliary Space: O(n)

An another Intuitive Solution can solve this problem with same complexity. 

The idea is to follow the algorithm used in Check if the parentheses is balanced or not. We follow this algorithm with a new condition when we find that the parentheses is not balanced.  This case arises when the stack is empty and we encounter a ‘ } ‘. In Check if the parentheses is balanced or not program we break the loop when we find that parentheses is not balanced but here we will reverse it to ‘ { ‘ and push it to the stack. While doing this, answer is incremented by 1.

Here, since we found a case of unbalanced expression the ‘ { ‘ must be changed in order to get a balanced expression. Also, changing this would be the most minimal way to get a balanced expression as it is a must condition to change it.

For example,  string = “}{{}}{}}” will be converted to “{{{}}{}}” and we get a balanced expression. There may arise a case where after doing this to the string we have some ‘{‘ left in the stack. For example,  string = “{}{{{{” will be converted to “{}{{{{” and there will be 4 ‘{‘ present in the stack which are not popped and are not balanced.

We can simply make it balanced by reversing the right half of the stack to ‘}’. Example: if stack has ‘ {{{{ ‘ left, we make it ‘ {{}} ‘ forming a balanced expression. Hence, answer gets updated by (stack size / 2). The case where the size of stack is odd, it is not possible to transform it to a balanced string.

Below is implementation of above idea: 

Output:

1

 Time Complexity: O(n)
 Auxiliary Space: O(n)

Another efficient solution solve the problem in O(1) i.e. constant space. Since the expression only contains one type of brackets, the idea is to maintain two variables to keep count of left bracket as well as right bracket as we did in Length of the longest valid substring. If the expression has balanced brackets, then we decrement left variable else we increment right variable. Then all we need to return is ceil(left/2) + ceil(right/2).

2

Time Complexity: O(n) 

Auxiliary Space: O(1)

Instead of maintaining two different variables for left brace and right brace, we can do it using a single temporary variable.

Traverse the array. For each ‘{‘ , increment the value of temp by 1 and for each ‘}’, if value of temp >0, then decrement the value of temp by 1 else, increment the value of result as well as temp by 1. At end, add half of the value of temp to the result.

Below is the implementation of above approach in C++.

2

Time Complexity: O(n)

Auxiliary Space: O(1)

Thanks to Utkarsh Trivedi for suggesting above approach.