The answer for the subproblem is, if the first letter of \(T'\) is 0, the number of times that 0 becomes 1 when inspecting \(T'\) in order from the first, or if the first letter of \(T'\) is 1, the number of times that 1 becomes 0 when inspecting \(T'\) in order from the first.
So, the problem can be solved this problem by a fast matrix exponentiation with the following five states:

• The first letter is 0 and the last letter is 0
• The first letter is 0 and the last letter is 1
• The first letter is 1 and the last letter is 1
• The first letter is 1 and the last letter is 0
• The answer for the subproblem for the characters currently inspected

Since ? can be seen as a superposition of 0 and 1, the matrix corresponding to ? is equal to the sum of the matrix corresponding to 0 and the matrix corresponding to 1.
By calculating the matrix for the second through the last character of \(T\), and applying it to the first character of \(T\), one can obtain the answer.
The total time complexity is \(O(|S| + \log K)\).

Sample Code (C++)