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

Find the Power of K-Size Subarrays II (LeetCode #3255) 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: Using a sliding window can significantly reduce the time complexity compared to checking each subarray individually.

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

The brute force approach for Find the Power of K-Size Subarrays II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible subarray of size k and determines if the 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 II?

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

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

Always explain your thought process clearly; it shows your understanding of the problem. Consider edge cases, such as when k equals 1 or when all elements are the same. Practice writing clean and efficient code, as interviewers often look for both correctness and performance.

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

Not properly handling the sliding window's left pointer, which can lead to incorrect frequency counts. Assuming that sorting is necessary for checking consecutive elements, which is inefficient.

#3255

Find the Power of K-Size Subarrays II

Medium

ArraySliding WindowSliding WindowHash Map

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)

Using a sliding window approach allows us to efficiently track the maximum element and check for consecutive elements without needing to sort each subarray. We can maintain a frequency map to ensure the elements are consecutive.

⚙️

Algorithm

3 steps

1Step 1: Initialize a frequency map to count occurrences of elements in the current window of size k.
2Step 2: Slide the window across the array, updating the frequency map as elements enter and exit the window.
3Step 3: Check if the window contains k unique elements and if the maximum element minus the minimum element equals k - 1. If true, return the maximum element; otherwise, return -1.

solution.py19 lines

1from collections import defaultdict
2
3def findPower(nums, k):
4    results = []
5    freq = defaultdict(int)
6    left = 0
7    for right in range(len(nums)):
8        freq[nums[right]] += 1
9        if right - left + 1 > k:
10            freq[nums[left]] -= 1
11            if freq[nums[left]] == 0:
12                del freq[nums[left]]
13            left += 1
14        if right - left + 1 == k:
15            if len(freq) == k and max(freq.keys()) - min(freq.keys()) == k - 1:
16                results.append(max(freq.keys()))
17            else:
18                results.append(-1)
19    return results

ℹ

Complexity note: The time complexity is O(n) because we process each element of the array once while maintaining a sliding window. The space complexity is O(n) due to the frequency map that can store up to n unique elements in the worst case.

1Using a sliding window can significantly reduce the time complexity compared to checking each subarray individually.
2Maintaining a frequency map allows for efficient checks on the uniqueness and consecutive nature of elements.

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