How do you solve Sliding Window Maximum on LeetCode?

Sliding Window Maximum (LeetCode #239) 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 deque allows us to efficiently manage the maximums in the current window.

What is the brute force solution for Sliding Window Maximum?

The brute force approach for Sliding Window Maximum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible window of size k and finding the maximum for each window. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sliding Window Maximum?

Sliding Window Maximum uses the following concepts: Array, Queue, Sliding Window, Heap (Priority Queue), Monotonic Queue. The recommended patterns to study are: Sliding Window, Deque, Monotonic Queue.

What are the interview tips for Sliding Window Maximum?

Explain your thought process clearly when discussing data structures. Practice writing code on a whiteboard or in an online editor to get comfortable with syntax. Be prepared to discuss time and space complexity for your solutions.

What are common mistakes when solving Sliding Window Maximum?

Not handling edge cases like empty arrays or k being larger than the array length. Forgetting to maintain the order of indices in the deque.

#239

Sliding Window Maximum

Hard

ArrayQueueSliding WindowHeap (Priority Queue)Monotonic QueueSliding WindowDequeMonotonic Queue

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses a deque (double-ended queue) to keep track of the indices of the maximum values in the current window. This allows us to efficiently add and remove elements as the window slides.

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Algorithm

4 steps

1Step 1: Initialize a deque to store indices of the elements in nums.
2Step 2: Loop through each element in nums, maintaining the deque to ensure it only contains indices of elements in the current window and in decreasing order of their values.
3Step 3: For each index, if the index is out of the current window, remove it from the front of the deque.
4Step 4: Append the maximum (nums[deque[0]]) to the result list once the first window is filled.

solution.py17 lines

1# Full working Python code
2from collections import deque
3
4def maxSlidingWindow(nums, k):
5    if not nums:
6        return []
7    max_values = []
8    dq = deque()
9    for i in range(len(nums)):
10        if dq and dq[0] < i - k + 1:
11            dq.popleft()
12        while dq and nums[dq[-1]] < nums[i]:
13            dq.pop()
14        dq.append(i)
15        if i >= k - 1:
16            max_values.append(nums[dq[0]])
17    return max_values

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Complexity note: The time complexity is O(n) because each element is added and removed from the deque at most once. The space complexity is O(n) due to the storage of indices in the deque.

1Using a deque allows us to efficiently manage the maximums in the current window.
2The sliding window technique is useful for problems involving contiguous subarrays.

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