Introduction

In this blog post, we will explore how to solve the LeetCode problem "3350" 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 nums of n integers, your task is to find the maximum value of k for which there exist two adjacent subarrays of length k each, such that both subarrays are strictly increasing. Specifically, check if there are two subarrays of length k starting at indices a and b (a < b), where: Both subarrays nums[a..a + k - 1] and nums[b..b + k - 1] are strictly increasing. The subarrays must be adjacent, meaning b = a + k. Return the maximum possible value of k. A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [2,5,7,8,9,2,3,4,3,1] Output: 3 Explanation: The subarray starting at index 2 is [7, 8, 9], which is strictly increasing. The subarray starting at index 5 is [2, 3, 4], which is also strictly increasing. These two subarrays are adjacent, and 3 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist. Example 2: Input: nums = [1,2,3,4,4,4,4,5,6,7] Output: 2 Explanation: The subarray starting at index 0 is [1, 2], which is strictly increasing. The subarray starting at index 2 is [3, 4], which is also strictly increasing. These two subarrays are adjacent, and 2 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist. Constraints: 2 <= nums.length <= 2 * 105 -109 <= nums[i] <= 109

Explanation

Here's the breakdown of the solution:

• Identify Increasing Subarrays: Iterate through the array and determine the lengths of all strictly increasing subarrays starting at each index.
• Check for Adjacent Pairs: Iterate through possible lengths k and check if there exist two adjacent strictly increasing subarrays of that length.

• Maximize k: Return the maximum k for which a pair of adjacent strictly increasing subarrays is found.

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

Code

    def max_adjacent_increasing_subarrays(nums):
    n = len(nums)
    max_k = 0

    # lengths[i] stores the length of the longest strictly increasing subarray
    # starting at index i
    lengths = [1] * n

    # Calculate the lengths of increasing subarrays
    for i in range(n - 2, -1, -1):
        if nums[i] < nums[i + 1]:
            lengths[i] = lengths[i + 1] + 1

    # Check for adjacent subarrays of length k
    for k in range(1, n // 2 + 1):
        for i in range(n - 2 * k + 1):
            if lengths[i] >= k and lengths[i + k] >= k:
                max_k = max(max_k, k)
                break  # Once we find a valid k, move to the next k
        else:
            continue # If no 'i' was breaked, no solution of this k exists
        continue # If the solution exists, continue with the next 'k'
        break # If the current k is not a solution, no solution exists for the rest of the 'k'

    return max_k