How do you solve Sliding Subarray Beauty on LeetCode?

Sliding Subarray Beauty (LeetCode #2653) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Maintaining a frequency count allows for efficient updates as the window slides.

What is the brute force solution for Sliding Subarray Beauty?

The brute force approach for Sliding Subarray Beauty is Brute Force, with O(n²) time complexity and O(k) space complexity. The brute force approach checks every possible subarray of size k, sorts the negative numbers, and retrieves the x-th smallest negative integer. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Sliding Subarray Beauty?

Sliding Subarray Beauty uses the following concepts: Array, Hash Table, Sliding Window. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Sliding Subarray Beauty?

Always clarify the problem constraints and edge cases before diving into coding. Think about how to optimize your solution after writing the brute force version. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Sliding Subarray Beauty?

Failing to handle the case where there are fewer than x negative numbers correctly. Not updating the frequency map properly when elements enter or exit the sliding window.

#2653

Sliding Subarray Beauty

Medium

ArrayHash TableSliding WindowHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

By maintaining a frequency count of negative numbers in the current window, we can efficiently determine the x-th smallest negative integer without sorting each time.

⚙️

Algorithm

4 steps

1Step 1: Use a frequency map to count negative numbers in the current window of size k.
2Step 2: For each new element added to the window, update the frequency map.
3Step 3: When moving the window, remove the element that is sliding out and update the frequency map accordingly.
4Step 4: To find the x-th smallest negative, iterate through the frequency map in sorted order.

solution.py21 lines

1from collections import defaultdict
2
3def beautyOfSubarrays(nums, k, x):
4    result = []
5    freq = defaultdict(int)
6    negatives = []
7
8    for i in range(len(nums)):
9        if nums[i] < 0:
10            freq[nums[i]] += 1
11            if freq[nums[i]] == 1:
12                negatives.append(nums[i])
13        if i >= k:
14            if nums[i - k] < 0:
15                freq[nums[i - k]] -= 1
16                if freq[nums[i - k]] == 0:
17                    negatives.remove(nums[i - k])
18        if i >= k - 1:
19            negatives.sort()
20            result.append(negatives[x - 1] if len(negatives) >= x else 0)
21    return result

ℹ

Complexity note: This complexity arises from maintaining the frequency map and sorting the negatives, which is efficient compared to the brute force approach.

1Maintaining a frequency count allows for efficient updates as the window slides.
2Sorting the negatives only when necessary reduces unnecessary computations.

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