Given an array of integers nums and an integer k. A continuous subarray is called nice if there are k odd numbers on it.
Return the number of nice sub-arrays.
Example 1:
Input: nums = [1,1,2,1,1], k = 3
Output: 2
Explanation: The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].
Example 2:
Input: nums = [2,4,6], k = 1
Output: 0
Explanation: There is no odd numbers in the array.
Example 3:
Input: nums = [2,2,2,1,2,2,1,2,2,2], k = 2
Output: 16
Constraints:
1 <= nums.length <= 50000
1 <= nums[i] <= 10^5
1 <= k <= nums.length
class Solution:
    def numberOfSubarrays(self, nums: List[int], k: int) -> int:
        # save all even subarray's length which between odds
        edge = []
        res = 0
        count = 0
        for i in nums:
            # odd
            if i % 2:
                # +1 because range from 0 to count when doing combination
                edge.append(count+1)
                count = 0
            # even
            else:
                count += 1
        edge.append(count+1)
        # no enough odd
        if len(edge)-1 < k:
            return 0
        else:
            # combination
            for i in range(len(edge)-k):
                res += edge[i] * edge[i+k]
            return res
class Solution:
    def numberOfSubarrays(self, nums: List[int], k: int) -> int:
        # save all even subarray's length which between odds
        edge = []
        res = 0
        count = 0
        for i in nums:
            # odd
            if i % 2:
                # +1 because range from 0 to count when doing combination
                edge.append(count+1)
                count = 0
            # even
            else:
                count += 1
        edge.append(count+1)
        # no enough odd
        if len(edge)-1 < k:
            return 0
        else:
            # combination
            for i in range(len(edge)-k):
                res += edge[i] * edge[i+k]
            return res