# Solving Leetcode Interviews in Seconds with AI: Adjacent Increasing Subarrays Detection II


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```