#3254

Medium

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Find the Power of K-Size Subarrays I

Given an array, find the power of all subarrays of size k using a sliding window approach.

Problem Statement

You are given an array nums of integers and a positive integer k. You need to compute the power of all subarrays of size k in nums. The power of an array is defined as the maximum value in the subarray minus the minimum value, provided that all elements in the subarray are distinct. If any elements are duplicated within the subarray, return -1 for that subarray.

You are required to solve the problem efficiently using a sliding window technique, where the subarrays are evaluated without recalculating the entire subarray repeatedly. Consider using a HashSet or other data structures to track distinct elements, ensuring the solution performs optimally for large inputs.

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 <= 500
1 <= nums[i] <= 105
1 <= k <= n

Solution Approach

Sliding Window with HashSet

Utilize a sliding window of size k that moves one element at a time. At each step, update the window by removing the first element of the previous window and adding the new element. Use a HashSet to track distinct elements in the current window, ensuring all elements are unique before computing the power.

Efficient Power Calculation

For each valid subarray, calculate the power as the difference between the maximum and minimum values. If the subarray contains duplicate values, return -1. Efficient calculation of the maximum and minimum values can be done by maintaining running values as the window slides.

Optimization with Time Complexity

The time complexity of this approach is O(n) because each element is added and removed from the window at most once. The space complexity is O(1) since only a few variables are required to track the current state of the window, aside from the HashSet.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

The solution runs in O(n) time, as each element in the array is processed exactly once due to the sliding window. The space complexity is O(1) because only a small fixed number of variables and a HashSet are used to maintain the state of the sliding window.

What Interviewers Usually Probe

Can the candidate efficiently implement the sliding window technique without recalculating values from scratch?
Does the candidate understand the importance of maintaining running state to optimize performance?
How well does the candidate manage the sliding window, especially with regard to duplicate elements?

Common Pitfalls or Variants

Common pitfalls

Failing to handle subarrays with duplicate elements correctly, leading to incorrect results for those subarrays.
Using a brute force method with nested loops that results in excessive time complexity, especially for larger inputs.
Not optimizing the update of the sliding window, causing unnecessary recalculations of maximum and minimum values.

Follow-up variants

Find the power of k-size subarrays where the elements must satisfy a different condition, such as being strictly increasing.
Adapt the solution to handle dynamic input sizes, where the value of k changes as the problem progresses.
Consider optimizing the solution by using a deque to keep track of the minimum and maximum values in the sliding window.

FAQ

How do I optimize the sliding window for the 'Find the Power of K-Size Subarrays I' problem?

You can optimize the sliding window by maintaining a running state with a HashSet to track distinct elements and by calculating the maximum and minimum values in O(1) time as the window slides.

What happens if there are duplicate elements in a subarray?

If any elements are duplicated within a subarray, the power of that subarray is -1.

What is the time complexity of solving 'Find the Power of K-Size Subarrays I'?

The time complexity is O(n), where n is the length of the array, since each element is added and removed from the sliding window at most once.

Can brute force be used for solving 'Find the Power of K-Size Subarrays I'?

While brute force can be used, it leads to a time complexity of O(n*k), which is inefficient for larger arrays. The sliding window approach is preferred for its O(n) time complexity.

What is the role of the HashSet in solving the problem?

The HashSet is used to track distinct elements in the sliding window and to ensure that the subarray is valid for power calculation.

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)]

Solution 2

#### TypeScript

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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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