<p>Given an array of integers <code>nums</code> and an integer <code>k</code>, return the number of <span data-keyword="subarray-nonempty">subarrays</span> of <code>nums</code> where the bitwise <code>AND</code> of the elements of the subarray equals <code>k</code>.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,1,1], k = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">6</span></p>

<p><strong>Explanation:</strong></p>

<p>All subarrays contain only 1&#39;s.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,1,2], k = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>Subarrays having an <code>AND</code> value of 1 are: <code>[<u><strong>1</strong></u>,1,2]</code>, <code>[1,<u><strong>1</strong></u>,2]</code>, <code>[<u><strong>1,1</strong></u>,2]</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,2,3], k = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p>Subarrays having an <code>AND</code> value of 2 are: <code>[1,<b><u>2</u></b>,3]</code>, <code>[1,<u><strong>2,3</strong></u>]</code>.</p>
</div>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i], k &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        scheme = [0] * 30
        pattern = []
        antipattern = []
        d = 0
        while k:
            if k % 2:
                pattern.append(d)
            else:
                antipattern.append(d)
            k //= 2
            d += 1
        for i in range(30):
            if i not in pattern and i not in antipattern:
                antipattern.append(i)
        
        def renew(x):
            d = 0
            while x:
                if x % 2:
                    if scheme[d] >= 0:
                        scheme[d] += 1
                    else:
                        scheme[d] = 1
                else:
                    if scheme[d] <= 0:
                        scheme[d] -= 1
                    else:
                        scheme[d] = -1
                x //= 2
                d += 1
            for i in range(d, 30):
                if scheme[i] <= 0:
                    scheme[i] -= 1
                else:
                    scheme[i] = -1

        def obsorbe(index):
            good = 10 ** 9
            bad = -10 ** 9
            bad_good = 0
            for i in pattern:
                good = min(good, scheme[i])
            for i in antipattern:
                bad = max(bad, scheme[i])
                bad_good = max(bad_good, scheme[i])
            if good < 0:
                return 0
            if pattern == []:
                return index + 1 - bad_good
            if bad < 0:
                return good
            return (good - bad) * int(good > bad)

        answer = 0
        for i in range(len(nums)):
            renew(nums[i])
            answer += obsorbe(i)
        return answer
4265