Introduction

In this blog post, we will explore how to solve the LeetCode problem "1438" 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

Given an array of integers nums and an integer limit, return the size of the longest non-empty subarray such that the absolute difference between any two elements of this subarray is less than or equal to limit. Example 1: Input: nums = [8,2,4,7], limit = 4 Output: 2 Explanation: All subarrays are: [8] with maximum absolute diff |8-8| = 0 <= 4. [8,2] with maximum absolute diff |8-2| = 6 > 4. [8,2,4] with maximum absolute diff |8-2| = 6 > 4. [8,2,4,7] with maximum absolute diff |8-2| = 6 > 4. [2] with maximum absolute diff |2-2| = 0 <= 4. [2,4] with maximum absolute diff |2-4| = 2 <= 4. [2,4,7] with maximum absolute diff |2-7| = 5 > 4. [4] with maximum absolute diff |4-4| = 0 <= 4. [4,7] with maximum absolute diff |4-7| = 3 <= 4. [7] with maximum absolute diff |7-7| = 0 <= 4. Therefore, the size of the longest subarray is 2. Example 2: Input: nums = [10,1,2,4,7,2], limit = 5 Output: 4 Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5. Example 3: Input: nums = [4,2,2,2,4,4,2,2], limit = 0 Output: 3 Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109 0 <= limit <= 109

Explanation

Here's the approach to solve this problem efficiently:

• Sliding Window: Use a sliding window approach to iterate through the array.
• Min/Max Deque: Maintain two double-ended queues (deques), one for tracking the minimum element within the current window and another for the maximum element. This allows us to find the min and max in O(1) time.

• Window Adjustment: Expand the window to the right. If the difference between the current window's maximum and minimum exceeds the limit, shrink the window from the left until the condition is met again.

• Runtime Complexity: O(n), where n is the length of the input array nums.

• Storage Complexity: O(n) in the worst-case scenario where the deques store all elements of nums.

Code

    from collections import deque

def longestSubarray(nums, limit):
    min_deque = deque()
    max_deque = deque()
    left = 0
    max_len = 0

    for right in range(len(nums)):
        # Maintain min_deque
        while min_deque and nums[min_deque[-1]] >= nums[right]:
            min_deque.pop()
        min_deque.append(right)

        # Maintain max_deque
        while max_deque and nums[max_deque[-1]] <= nums[right]:
            max_deque.pop()
        max_deque.append(right)

        # Shrink window if necessary
        while nums[max_deque[0]] - nums[min_deque[0]] > limit:
            if left == min_deque[0]:
                min_deque.popleft()
            if left == max_deque[0]:
                max_deque.popleft()
            left += 1

        # Update max_len
        max_len = max(max_len, right - left + 1)

    return max_len