<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 numberOfSubarrays(self, nums: List[int], k: int) -> int:
        prefix = []
        for i in range(len(nums)):
            if nums[i]%2:
                nums[i]=1
            else:
                nums[i]=0
        total = 0
        for i in range(len(nums)):
            total+=nums[i]
            prefix.append(total)
        n = len(prefix)
        freq = Counter(prefix)
        count=0
        count+=prefix.count(k)
        prefix = list(set(prefix))
        i=0
        j=1
        while j<len(prefix):
            diff = prefix[j]-prefix[i]
            if diff>k:
                i+=1
            elif diff<k:
                j+=1
            else:
                count+= freq[prefix[i]]*freq[prefix[j]]
                i+=1
                j+=1
        return count


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