# Solving Leetcode Interviews in Seconds with AI: The Number of Beautiful Subsets


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2597" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given an array nums of positive integers and a positive integer k. A subset of nums is beautiful if it does not contain two integers with an absolute difference equal to k. Return the number of non-empty beautiful subsets of the array nums. A subset of nums is an array that can be obtained by deleting some (possibly none) elements from nums. Two subsets are different if and only if the chosen indices to delete are different.   Example 1:  Input: nums = [2,4,6], k = 2 Output: 4 Explanation: 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].  Example 2:  Input: nums = [1], k = 1 Output: 1 Explanation: The beautiful subset of the array nums is [1]. It can be proved that there is only 1 beautiful subset in the array [1].    Constraints:  1 <= nums.length <= 18 1 <= nums[i], k <= 1000  

	# Explanation
	Here's the solution:

*   **Divide and Conquer with Bitmasking:** Split the input array `nums` into two halves. Generate all possible subsets for each half using bitmasking.
*   **Check for Beautiful Subsets:** For each subset from the first half, check which subsets from the second half can be combined with it to form a beautiful subset (no absolute difference of `k`).
*   **Count Valid Combinations:** Sum the number of valid beautiful subsets formed by combining subsets from both halves.

*   **Time Complexity:** O(2<sup>n/2</sup> * 2<sup>n/2</sup>) = O(4<sup>n/2</sup>) which simplifies to roughly O(2<sup>n</sup>) since we can consider half the input length in the exponent. Specifically, the complexity is determined by iterating through all subsets of both halves and checking compatibility, which results in this exponential behavior, even though the input is split.
    **Space Complexity:** O(2<sup>n/2</sup>), which stores the subsets generated from each half.

	
	# Code
	```python
	def beautifulSubsets(nums, k):
    n = len(nums)

    def generate_subsets(arr):
        subsets = []
        for i in range(1 << len(arr)):
            subset = []
            for j in range(len(arr)):
                if (i >> j) & 1:
                    subset.append(arr[j])
            subsets.append(subset)
        return subsets

    first_half = nums[:n // 2]
    second_half = nums[n // 2:]

    subsets1 = generate_subsets(first_half)
    subsets2 = generate_subsets(second_half)

    count = -1  # Subtract empty set
    for sub1 in subsets1:
        for sub2 in subsets2:
            combined_set = sub1 + sub2
            if not combined_set:
                continue
            is_beautiful = True
            for i in range(len(combined_set)):
                for j in range(i + 1, len(combined_set)):
                    if abs(combined_set[i] - combined_set[j]) == k:
                        is_beautiful = False
                        break
                if not is_beautiful:
                    break
            if is_beautiful:
                count += 1

    return count
	```