🌟 Discovering Beautiful Subsets: Unleash the Power of Backtracking! 💫
Intuition\nThe problem is to find subsets such that no two elements in the subset have an absolute difference equal to k. This problem naturally lends itself to
# Intuition\nThe problem is to find subsets such that no two elements in the subset have an absolute difference equal to k. This problem naturally lends itself to a backtracking solution where we explore all possible subsets and only include those that meet the criteria.\n# Approach\nBacktracking Setup: Use a recursive backtracking function to explore all subsets of the given array.\nSubset Validity: For each subset being considered, check if adding a new element maintains the "beautiful" property (i.e., no two elements differ by k).\nRecursive Exploration: Recursively add elements to the current subset and count all valid subsets.\nBase and Recursive Case: The base case is when we reach the end of the array. The recursive case involves including the current element and moving to the next one, ensuring the new element doesn\'t violate the subset\'s property.\n# Complexity\n- Time complexity:\nThe time complexity is O(2^n\u22C5n). In the worst case, we generate all 2^n subsets and for each subset, we might check each of the elements(up to n elements) to ensure it maintains the "beautiful" property.\n- Space complexity:The space complexity is O(n) for the recursion stack and subset storage, where n is the length of the input array.\n# Code\n```\n/**\n * @param {number[]} nums\n * @param {number} k\n * @return {number}\n */\nvar beautifulSubsets = function(nums, k) {\n    const n = nums.length;\n    let result = 0;\n\n    const backtrack = (start, subset) => {\n        if (subset.length > 0) {\n            result++;\n        }\n        for (let i = start; i < n; i++) {\n            let valid = true;\n            for (let num of subset) {\n                if (Math.abs(num - nums[i]) === k) {\n                    valid = false;\n                    break;\n                }\n            }\n            if (valid) {\n                subset.push(nums[i]);\n                backtrack(i + 1, subset);\n                subset.pop();\n            }\n        }\n    };\n\n    backtrack(0, []);\n    return result;\n};\n\n```
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