Brute Force Solution

A naive approach would be to iterate through the array with a sliding window of size k. For each window, find the maximum element. This involves iterating k times for each window.

Code (Python):

def max_sliding_window_brute_force(nums, k):
    if not nums:
        return []
    
    result = []
    for i in range(len(nums) - k + 1):
        window = nums[i:i+k]
        result.append(max(window))
    return result

Big O Analysis:

• Time Complexity: O(n*k), where n is the length of the input array nums and k is the window size. This is because for each of the n-k+1 windows, we iterate through k elements to find the maximum.
• Space Complexity: O(n-k+1) in the worst case, where n is the length of the input array nums and k is the size of the sliding window, to store the output array. Can be O(1) if the result is computed in place.

Optimal Solution: Using a Deque

A more efficient solution utilizes a double-ended queue (deque) to maintain a decreasing order of indices of potentially maximum elements within the window.

Algorithm:

1. Initialize a deque.
2. Iterate through the array nums.
3. For each element:
   • Remove elements from the rear of the deque that are smaller than the current element. This ensures the deque maintains a decreasing order.
   • Add the current element's index to the rear of the deque.
   • Remove elements from the front of the deque that are outside the current window (i.e., their index is less than i - k + 1).
   • If the window is full (i.e., i >= k - 1), the front of the deque contains the index of the maximum element in the current window. Add this element to the result.

Code (Python):

from collections import deque

def max_sliding_window(nums, k):
    if not nums:
        return []

    d = deque()
    result = []

    for i in range(len(nums)):
        # Remove elements out of the window
        while d and d[0] <= i - k:
            d.popleft()

        # Remove smaller elements in deque than current element
        while d and nums[d[-1]] <= nums[i]:
            d.pop()

        d.append(i)

        # Add max to results
        if i >= k - 1:
            result.append(nums[d[0]])

    return result

Big O Analysis:

• Time Complexity: O(n), where n is the length of the input array nums. Each element is added and removed from the deque at most once.
• Space Complexity: O(k), where k is the window size, for storing the deque.

Edge Cases:

• Empty input array: Return an empty array.
• k is 1: Return the original array.
• k is equal to the length of nums: Return an array with a single element, which is the maximum of nums.
• k is greater than the length of nums: This violates the constraint. Should throw an IllegalArgumentException.