The Number of Beautiful Subsets
<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;= 18</code></li>
	<li><code>1 &lt;= nums[i], k &lt;= 1000</code></li>
</ul>
Array; Hash Table; Math; Dynamic Programming; Backtracking; Combinatorics; Sorting
class Solution {
    private int[] nums;
    private int[] cnt = new int[1010];
    private int ans = -1;
    private int k;

    public int beautifulSubsets(int[] nums, int k) {
        this.k = k;
        this.nums = nums;
        dfs(0);
        return ans;
    }

    private void dfs(int i) {
        if (i >= nums.length) {
            ++ans;
            return;
        }
        dfs(i + 1);
        boolean ok1 = nums[i] + k >= cnt.length || cnt[nums[i] + k] == 0;
        boolean ok2 = nums[i] - k < 0 || cnt[nums[i] - k] == 0;
        if (ok1 && ok2) {
            ++cnt[nums[i]];
            dfs(i + 1);
            --cnt[nums[i]];
        }
    }
}