<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 normalise_array(self, nums: List[int]):
        for i, num in enumerate(nums):
            if num % 2:
                nums[i] = 1
            else:
                nums[i] = 0

        return nums

    def numberOfSubarrays(self, nums: List[int], k: int) -> int:
        """
        # HashMap
        curr_sum = 0
        subarrays = 0
        prefix_sum = {curr_sum: 1}

        nums = self.normalise_array(nums)

        for i in range(len(nums)):
            curr_sum += nums[i]
            # Find subarrays with sum k ending at i
            if curr_sum - k in prefix_sum:
                subarrays += prefix_sum[curr_sum - k]
            # Increment the current prefix sum in hashmap
            prefix_sum[curr_sum] = prefix_sum.get(curr_sum, 0) + 1

        return subarrays
        """

        odd_indices = deque()
        subarrays = 0
        last_popped = -1
        initial_gap = -1

        nums = self.normalise_array(nums)

        for i in range(len(nums)):
            # If element is odd, append its index to the deque.
            if nums[i]:
                odd_indices.append(i)
            # If the number of odd numbers exceeds k, remove the first odd index.
            if len(odd_indices) > k:
                last_popped = odd_indices.popleft()
            # If there are exactly k odd numbers, add the number of even numbers
            # in the beginning of the subarray to the result.
            if len(odd_indices) == k:
                initial_gap = odd_indices[0] - last_popped
                subarrays += initial_gap

        return subarrays
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