Count Number of Maximum Bitwise-OR Subsets
<p>Given an integer array <code>nums</code>, find the <strong>maximum</strong> possible <strong>bitwise OR</strong> of a subset of <code>nums</code> and return <em>the <strong>number of different non-empty subsets</strong> with the maximum bitwise OR</em>.</p>

<p>An array <code>a</code> is a <strong>subset</strong> of an array <code>b</code> if <code>a</code> can be obtained from <code>b</code> by deleting some (possibly zero) elements of <code>b</code>. Two subsets are considered <strong>different</strong> if the indices of the elements chosen are different.</p>

<p>The bitwise OR of an array <code>a</code> is equal to <code>a[0] <strong>OR</strong> a[1] <strong>OR</strong> ... <strong>OR</strong> a[a.length - 1]</code> (<strong>0-indexed</strong>).</p>

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

<pre>
<strong>Input:</strong> nums = [3,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The maximum possible bitwise OR of a subset is 3. There are 2 subsets with a bitwise OR of 3:
- [3]
- [3,1]
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,2,2]
<strong>Output:</strong> 7
<strong>Explanation:</strong> All non-empty subsets of [2,2,2] have a bitwise OR of 2. There are 2<sup>3</sup> - 1 = 7 total subsets.
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,2,1,5]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The maximum possible bitwise OR of a subset is 7. There are 6 subsets with a bitwise OR of 7:
- [3,5]
- [3,1,5]
- [3,2,5]
- [3,2,1,5]
- [2,5]
- [2,1,5]</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 16</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Bit Manipulation; Array; Backtracking; Enumeration
class Solution:
    def countMaxOrSubsets(self, nums: List[int]) -> int:
        def dfs(i, t):
            nonlocal ans, mx
            if i == len(nums):
                if t == mx:
                    ans += 1
                return
            dfs(i + 1, t)
            dfs(i + 1, t | nums[i])

        ans = 0
        mx = reduce(lambda x, y: x | y, nums)
        dfs(0, 0)
        return ans
