"Efficiently Finding Beautiful Subsets Using Depth-First Search and Backtracking"
Intuition\nWhen solving this problem, my initial thought was to use a depth-first search (DFS) to explore all possible subsets of the given array. The key chall
## Intuition\nWhen solving this problem, my initial thought was to use a depth-first search (DFS) to explore all possible subsets of the given array. The key challenge is to ensure that for any subset, the difference between any two elements is not equal to `k`. To efficiently manage and check these conditions during the subset construction, a map can be used to track the inclusion of elements and their constraints.\n\n## Approach\nThe solution involves a recursive DFS function that explores all subsets of the given array `nums`. Here\'s a step-by-step breakdown of the approach:\n\n1. **Base Case:** If we have processed all elements in the array, return 1 to count the subset (including the empty subset).\n\n2. **Recursive Case:**\n    - **Include the current element:** Before including an element, check if it does not violate the condition with the existing elements in the subset (i.e., no element should differ from the current element by `k`). If it can be included:\n        - Add the element to the map and recursively call the DFS for the next element.\n        - After the recursive call, backtrack by removing the element from the map to restore the state for exploring other subsets.\n    - **Exclude the current element:** Recursively call the DFS for the next element without including the current element.\n\n3. **Combine Results:** The total number of beautiful subsets from the current state is the sum of subsets formed by including and excluding the current element.\n\n4. **Final Adjustment:** Subtract 1 from the total count to exclude the empty subset.\n\nBy using a map to keep track of the elements included in the current subset and their constraints, we can efficiently manage the conditions and explore all valid subsets.\n\n## Complexity\n- **Time complexity:** The time complexity is \\(O(2^n)\\), where \\(n\\) is the number of elements in the array. This is because, in the worst case, we explore all possible subsets.\n- **Space complexity:** The space complexity is \\(O(n)\\) due to the recursive call stack and the map used for tracking included elements and their counts.\n\n\n\n# Code\n```\nclass Solution {\n    public int beautifulSubsets(int[] nums, int k) {\n        Map<Integer, Integer> map = new HashMap<>();\n        // Starting the DFS traversal from index 0\n        int ans = dfs(nums, 0, k, map);\n        // Subtract 1 to exclude the empty subset\n        return ans - 1;\n    }\n    \n    public int dfs(int[] nums, int idx, int k, Map<Integer, Integer> map) {\n        // Base case: If we have processed all elements\n        if (idx == nums.length) return 1;\n        \n        int taken = 0;\n        // Check if the current element can be taken\n        if (!map.containsKey(nums[idx] - k) && !map.containsKey(nums[idx] + k)) {\n            // Include the current element\n            map.put(nums[idx], map.getOrDefault(nums[idx], 0) + 1);\n            taken = dfs(nums, idx + 1, k, map);\n            // Backtrack: Remove the current element\n            map.put(nums[idx], map.get(nums[idx]) - 1);\n            if (map.get(nums[idx]) == 0) map.remove(nums[idx]);\n        }\n        \n        // Exclude the current element\n        int notTaken = dfs(nums, idx + 1, k, map);\n        \n        // Total beautiful subsets from this state\n        return taken + notTaken;\n    }\n}\n```
['Java']