left and right, to represent the window's boundaries, and a variable max_size to keep track of the maximum window size found.right pointer over the answerKey string from left to right.
answerKey[right]).k, increment the left pointer and decrement the counter for the answer that is no longer in the window.max_size to be the maximum of its current value and the current window size (right - left + 1).max_size as the result.