This problem can be transformed into finding the number of subarrays with sum k by replacing odd numbers with 1 and even numbers with 0. However, a sliding window "At Most K" approach is more efficient. Exactly K = AtMost(K) - AtMost(K-1).

1. Define at_most(k):
   • Use sliding window to find total subarrays with at most k odd numbers.
   • For every r, expand until odds > k. Shrink l.
   • count += (r - l + 1).
2. Return at_most(k) - at_most(k - 1).

• Time Complexity: O(N).
• Space Complexity: O(1).

def count_nice_subarrays(nums, k):
    def at_most(goal):
        if goal < 0: return 0
        res = 0
        l = 0
        odds = 0
        for r in range(len(nums)):
            if nums[r] % 2 != 0:
                odds += 1
            while odds > goal:
                if nums[l] % 2 != 0:
                    odds -= 1
                l += 1
            res += (r - l + 1)
        return res
        
    return at_most(k) - at_most(k - 1)