Very simple Java backtracking solution
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    int count = 0;\n    public int beautifulSubsets(int[] nums, int k) {\n        Arrays.sort(nums);\n        backtrack(new HashSet<>(), 0, nums, k);\n        return count - 1;\n    }\n    public void backtrack(Set<Integer> current, int i, int[] nums, int k) {\n        count++;\n        for (int j = i; j < nums.length; j++) {\n            if (!current.contains(nums[j] - k)) {\n                current.add(nums[j]);\n                backtrack(current, j + 1, nums, k);\n                current.remove(Integer.valueOf(nums[j]));\n            }\n        }\n    }\n}\n```
['Java']