How do you solve The Number of Beautiful Subsets on LeetCode?

The Number of Beautiful Subsets (LeetCode #2597) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Sorting helps in managing the conditions for beautiful subsets.

What is the brute force solution for The Number of Beautiful Subsets?

The brute force approach for The Number of Beautiful Subsets is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsets of the given array and checking each subset to see if it meets the beautiful condition. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for The Number of Beautiful Subsets?

Always clarify the definition of a beautiful subset before starting. Think about edge cases, such as when k is larger than any element in nums. Practice backtracking problems to get comfortable with recursive solutions.

What are common mistakes when solving The Number of Beautiful Subsets?

Not considering the empty subset as invalid. Failing to sort the array before processing.

#2597

The Number of Beautiful Subsets

Medium

ArrayHash TableMathDynamic ProgrammingBacktrackingSortingCombinatoricsHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a backtracking approach with a frequency array to track the counts of elements, allowing us to efficiently skip adding elements that would violate the beautiful condition. This reduces unnecessary checks and speeds up the process.

⚙️

Algorithm

3 steps

1Step 1: Sort the input array nums.
2Step 2: Create a frequency array to count occurrences of each number.
3Step 3: Use backtracking to explore subsets, adding elements only if they do not create a conflict with the existing subset.

solution.py24 lines

1# Full working Python code
2from collections import defaultdict
3
4def beautiful_subsets(nums, k):
5    nums.sort()
6    count = 0
7    freq = defaultdict(int)
8    def backtrack(index):
9        nonlocal count
10        if index == len(nums):
11            return
12        # Option to not include nums[index]
13        backtrack(index + 1)
14        # Option to include nums[index]
15        if freq[nums[index] - k] == 0:
16            freq[nums[index]] += 1
17            count += 1  # Count this subset
18            backtrack(index + 1)
19            freq[nums[index]] -= 1
20    backtrack(0)
21    return count
22
23# Example usage
24print(beautiful_subsets([2, 4, 6], 2))  # Output: 4

ℹ

Complexity note: This complexity arises because we efficiently track the frequency of elements, allowing us to skip checks for subsets that would violate the beautiful condition, thus reducing the overall time taken.

1Sorting helps in managing the conditions for beautiful subsets.
2Using a frequency array allows for quick checks on whether to include an element.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.