exactly K odd numbers. This works because: countExactlyK = countAtMost(K) - countAtMost(K - 1).

Use a helper function to count the number of subarrays with at most a given number of odd numbers.
Initialize two pointers to define the sliding window.
Expand the window by moving the right pointer.
If the number of odd numbers in the window exceeds the allowed count, move the left pointer to shrink the window.
For each valid window, the number of subarrays ending at the current index is (right - left + 1).
Repeat for K and K - 1, then return their difference.