Introduction

In this blog post, we will explore how to solve the LeetCode problem "239" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an array of integers nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Return the max sliding window. Example 1: Input: nums = [1,3,-1,-3,5,3,6,7], k = 3 Output: [3,3,5,5,6,7] Explanation: Window position Max --------------- ----- [1 3 -1] -3 5 3 6 7 3 1 [3 -1 -3] 5 3 6 7 3 1 3 [-1 -3 5] 3 6 7 5 1 3 -1 [-3 5 3] 6 7 5 1 3 -1 -3 [5 3 6] 7 6 1 3 -1 -3 5 [3 6 7] 7 Example 2: Input: nums = [1], k = 1 Output: [1] Constraints: 1 <= nums.length <= 105 -104 <= nums[i] <= 104 1 <= k <= nums.length

Explanation

Here's an efficient solution to find the maximum sliding window, along with explanations:

• Key Approach: Use a double-ended queue (deque) to store indices of potentially maximum elements within the window. The deque maintains a decreasing order, ensuring the front always holds the index of the maximum element in the current window.
• Optimization: Before adding a new element's index, remove indices from the back of the deque that correspond to elements smaller than the current element. Also, remove indices from the front if they are outside the current window. This keeps the deque relevant and efficient.
• Runtime & Space Complexity: O(n) runtime, O(k) space.

Code

    from collections import deque

def maxSlidingWindow(nums, k):
    """
    Finds the maximum of each sliding window of size k in an array.

    Args:
      nums: A list of integers.
      k: The size of the sliding window.

    Returns:
      A list of the maximums of each sliding window.
    """

    if not nums:
        return []

    if k == 1:
        return nums

    result = []
    dq = deque()  # Store indices

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

        # Remove smaller elements from the back
        while dq and nums[dq[-1]] <= nums[i]:
            dq.pop()

        dq.append(i)

        # Add the maximum to the result
        if i >= k - 1:
            result.append(nums[dq[0]])

    return result