<p>You are given an array <code>nums</code> of positive integers and a <strong>positive</strong> integer <code>k</code>.</p>

<p>A subset of <code>nums</code> is <strong>beautiful</strong> if it does not contain two integers with an absolute difference equal to <code>k</code>.</p>

<p>Return <em>the number of <strong>non-empty beautiful </strong>subsets of the array</em> <code>nums</code>.</p>

<p>A <strong>subset</strong> of <code>nums</code> is an array that can be obtained by deleting some (possibly none) elements from <code>nums</code>. Two subsets are different if and only if the chosen indices to delete are different.</p>

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

<pre>
<strong>Input:</strong> nums = [2,4,6], k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> The beautiful subsets of the array nums are: [2], [4], [6], [2, 6].
It can be proved that there are only 4 beautiful subsets in the array [2,4,6].
</pre>

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

<pre>
<strong>Input:</strong> nums = [1], k = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> The beautiful subset of the array nums is [1].
It can be proved that there is only 1 beautiful subset in the array [1].
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 20</code></li>
	<li><code>1 &lt;= nums[i], k &lt;= 1000</code></li>
</ul>

class Solution:
    def beautifulSubsets(self, nums: List[int], k: int) -> int:
        n = len(nums)
        blacklist = [0] * 1001
        self.count = 0

        def dfs(cur_idx, path):
            if cur_idx == n:
                self.count += 1
                return
            dfs(cur_idx+1, path)
            # when adding, we need to chekck if its allowed
            if not blacklist[nums[cur_idx]]:
                # when updating the blacklist, check if the blacklisted number is in the allowed range
                if nums[cur_idx] + k in range(0,1001):
                    blacklist[nums[cur_idx] + k] += 1
                if nums[cur_idx] - k in range(0,1001):
                    blacklist[nums[cur_idx] - k] += 1
                dfs(cur_idx+1, path + [nums[cur_idx]])
                # when updating the blacklist, check if the blacklisted number is in the allowed range
                if nums[cur_idx] + k in range(0,1001):
                    blacklist[nums[cur_idx] + k] -= 1
                if nums[cur_idx] - k in range(0,1001):
                    blacklist[nums[cur_idx] - k] -= 1
        dfs(0, [])
        return self.count - 1 # -1 to exclude []
17100