<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:
        def atmost_k(k):
            n = len(nums)
            l = 0
            res = 0
            # memo = [[0] * 32 for _ in range(n)] 
            curr = [0] * 32
            for r in range(n):
                temp = nums[r]
                num = 0
                for i in range(31, -1, -1):
                    curr[i] += temp & 1
                    # memo[r][i] = temp & 1
                    temp >>= 1
                    if curr[i] == r - l + 1:
                        num |= 1 << (31 - i)
                    if temp == 0:
                        break
                while l <= r and num <= k:
                    res += n - r
                    new = 0
                    temp = nums[l]
                    for i in range(31, -1, -1):
                        curr[i] -= temp & 1
                        temp >>= 1
                        if r > l and curr[i] == r - l:
                            new |= 1 << (31 - i)
                        if nums[l] == 0:
                            break
                    num = new
                    l += 1
            return res
        ans = atmost_k(k) - atmost_k(k - 1)
        # print(ans)  
        return ans

4453