How do you solve Find the Power of K-Size Subarrays I on LeetCode?

Find the Power of K-Size Subarrays I (LeetCode #3254) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(k) space complexity. Key insight: Using a sliding window allows for efficient checking of subarrays without redundant calculations.

What is the brute force solution for Find the Power of K-Size Subarrays I?

The brute force approach for Find the Power of K-Size Subarrays I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each subarray of size k to see if its elements are consecutive and sorted. If they are, we return the maximum element; otherwise, we return -1. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Power of K-Size Subarrays I?

Find the Power of K-Size Subarrays I uses the following concepts: Array, Sliding Window. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Power of K-Size Subarrays I?

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

What are common mistakes when solving Find the Power of K-Size Subarrays I?

Not checking if elements are unique, leading to incorrect results. Misunderstanding the conditions for consecutive elements.

#3254

Find the Power of K-Size Subarrays I

Medium

ArraySliding WindowHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(k)

The optimal solution uses a sliding window approach to efficiently check each subarray of size k. We maintain a set to track the elements and their count, ensuring they are unique and consecutive.

⚙️

Algorithm

3 steps

1Step 1: Initialize a sliding window of size k and a HashSet to track unique elements.
2Step 2: As you slide the window, add the new element and remove the old one, updating the HashSet.
3Step 3: Check if the maximum element minus the minimum element equals k-1 and if the size of the HashSet equals k. If true, store the maximum element; otherwise, store -1.

solution.py13 lines

1# Full working Python code
2results = []
3window = set()
4for i in range(len(nums)):
5    window.add(nums[i])
6    if i >= k:
7        window.remove(nums[i - k])
8    if i >= k - 1:
9        if max(window) - min(window) == k - 1 and len(window) == k:
10            results.append(max(window))
11        else:
12            results.append(-1)
13

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once, and the operations on the HashSet (add, remove, max, min) are average O(1). The space complexity is O(k) due to the storage of elements in the HashSet.

1Using a sliding window allows for efficient checking of subarrays without redundant calculations.
2Maintaining a set helps track unique elements and their relationships easily.

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