<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:
        
        """
        [1,1,2,1,1]
        
        [1,2,2,3,4]

        [4,3,2,2,1]
        """
        ans = 0
        pref_dict = defaultdict(list)
        pref_sum = [0]
        pref_dict[0].append(0)
        for indx, curr_n in enumerate(nums[::-1]):
            if curr_n % 2 == 1:
                pref_sum.append(pref_sum[-1] + 1)
            else:
                pref_sum.append(pref_sum[-1])
            curr_sum = pref_sum[-1]
            if curr_sum >= k:
                ans += len(pref_dict[curr_sum - k])
            # we need 3 
            pref_dict[pref_sum[-1]].append(index)
            # print(pref_sum)
            # print(f'{indx}/{curr_n}/{curr_sum}')
            # print(ans)
        # print(pref_sum)
        # print(pref_dict)
        # ans = 0 
        # for curr_sum in pref_sum[1:]:
        #     ans += len(pref_dict[k - curr_sum])

        return ans


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