<p>Given an array of integers <code>nums</code> and an integer <code>k</code>. A continuous subarray is called <strong>nice</strong> if there are <code>k</code> odd numbers on it.</p>

<p>Return <em>the number of <strong>nice</strong> sub-arrays</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,2,1,1], k = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,4,6], k = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no odd numbers in the array.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2,2,1,2,2,1,2,2,2], k = 2
<strong>Output:</strong> 16
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 50000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10^5</code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

class Solution:
    def count_sub(self, nums, k):
        n = len(nums)
        left, right = 0, 0
        track = {}
        res = 0
        count = 0

        while right < n:
            track[nums[right]] = track.get(nums[right], 0) + 1
            if nums[right] % 2 != 0:
                count += 1
            while count > k and left <= right:
                track[nums[left]] -= 1
                if nums[left] % 2 != 0:
                    count -= 1
                left += 1
            if count <= k:
                res += right - left + 1
            right += 1
        return res
    def numberOfSubarrays(self, nums: List[int], k: int) -> int:
        n = len(nums)
        left, right = 0, 0
        track = {}
        count = 0

        return self.count_sub(nums, k) - self.count_sub(nums, k - 1)
24500