<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 __init__(self):
        self.answer = 0

    def is_beautiful(self, count, k):
        for num in count:
            if num + k in count or num - k in count:
                return False
        return True

    def backtrack(self, nums, k, curr_count, idx):
        """
        :param curr_count: dict of the count of each number we have chosen BEFORE idx
        :param idx: the index of the number which we are going to insert into count
        """
        # insert this new number into the curr_count dictionary
        if nums[idx] in curr_count:
            curr_count[nums[idx]] += 1
        else:
            curr_count[nums[idx]] = 1
        
        # try adding each remaining element into the subarray as well
        for j in range(idx + 1, len(nums)):
            self.backtrack(nums, k, {**curr_count}, j)
        # given the current subarray, check if it is beautiful
        if self.is_beautiful(curr_count, k):
            self.answer += 1

    def beautifulSubsets(self, nums: List[int], k: int) -> int:
        for i in range(len(nums)):
            self.backtrack(nums, k, {}, i)
        return self.answer
16300