#3255

Medium

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Find the Power of K-Size Subarrays II

Compute the power of all k-size subarrays in an array using sliding window with running state updates efficiently.

Problem Statement

You are given an integer array nums of length n and a positive integer k. The power of an array is defined based on the longest consecutive sequence of sorted elements within each k-size subarray. Your task is to compute the power for every subarray of size k in nums efficiently using sliding window updates.

Return an array of length n - k + 1 where each element represents the power of the corresponding k-size subarray. Optimize your solution to handle large n up to 105 and ensure proper handling of repeated elements or sequences that break consecutiveness.

Examples

Example 1

Input: nums = [1,2,3,4,3,2,5], k = 3

Output: [3,4,-1,-1,-1]

There are 5 subarrays of nums of size 3:

Example 2

Input: nums = [2,2,2,2,2], k = 4

Output: [-1,-1]

Example details omitted.

Example 3

Input: nums = [3,2,3,2,3,2], k = 2

Output: [-1,3,-1,3,-1]

Example details omitted.

Constraints

1 <= n == nums.length <= 105
1 <= nums[i] <= 106
1 <= k <= n

Solution Approach

Sliding Window Initialization

Start by computing the initial window of size k, determining the power based on consecutive increasing sequences. Maintain auxiliary state variables that track positions and counts needed to compute the power efficiently as the window slides.

Window Update and State Maintenance

As you slide the window forward, update the running state by removing the outgoing element and adding the incoming element. Recalculate power only using the changes instead of re-evaluating the entire window, ensuring O(n) traversal overall.

Result Collection and Edge Handling

Store the computed power for each window in the result array. Handle edge cases such as repeated elements, windows where the consecutive sequence breaks, or k equals 1, ensuring correctness for all input constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n) because each element is processed once while updating the sliding window. Space complexity is O(k) or O(n) depending on whether auxiliary arrays are maintained to track sequences.

What Interviewers Usually Probe

Expect optimized O(n) solution using sliding window with running state updates.
Look for careful handling of sequences that break consecutiveness within windows.
Assess if edge cases with repeated numbers or k = 1 are correctly handled.

Common Pitfalls or Variants

Common pitfalls

Recomputing the entire window's power from scratch on each slide leading to O(n*k) time.
Failing to update state correctly when the outgoing element affects the consecutive sequence.
Incorrectly handling repeated numbers that disrupt the consecutive order in a subarray.

Follow-up variants

Compute the sum of powers instead of individual subarray powers.
Find the maximum power across all k-size subarrays instead of returning the array.
Extend the problem to variable-size windows and track the longest consecutive sequence for each size.

FAQ

What is the core pattern used in Find the Power of K-Size Subarrays II?

The problem relies on a sliding window with running state updates to compute subarray powers without recomputing each window.

Can this algorithm handle repeated elements correctly?

Yes, careful state tracking ensures that repeated numbers disrupting sequences are handled while computing the power.

What is the time complexity of the recommended solution?

The solution runs in O(n) time because each element is added and removed from the sliding window once while maintaining the state.

How should edge cases like k = 1 be handled?

For k = 1, each element forms a subarray of size 1, so the power is determined directly from the element itself without window comparison.

Is it necessary to store auxiliary state for sequences?

Yes, auxiliary state is required to track consecutive increasing sequences efficiently as the window slides and prevents recalculating full subarrays.

terminal

Solution

Solution 1: Recursion

We define an array $f$, where $f[i]$ represents the length of the continuous increasing subsequence ending at the $i$-th element. Initially, $f[i] = 1$.

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class Solution:
    def resultsArray(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        f = [1] * n
        for i in range(1, n):
            if nums[i] == nums[i - 1] + 1:
                f[i] = f[i - 1] + 1
        return [nums[i] if f[i] >= k else -1 for i in range(k - 1, n)]

Continue Practicing

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