# Solving Leetcode Interviews in Seconds with AI: Maximum Length of Repeated Subarray


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "718" 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 two integer arrays nums1 and nums2, return the maximum length of a subarray that appears in both arrays.   Example 1:  Input: nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7] Output: 3 Explanation: The repeated subarray with maximum length is [3,2,1].  Example 2:  Input: nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0] Output: 5 Explanation: The repeated subarray with maximum length is [0,0,0,0,0].    Constraints:  1 <= nums1.length, nums2.length <= 1000 0 <= nums1[i], nums2[i] <= 100  

	# Explanation
	Here's an efficient solution to find the maximum length of a common subarray between two arrays:

*   **Dynamic Programming:** Utilize a 2D DP table where `dp[i][j]` stores the length of the longest common subarray ending at `nums1[i-1]` and `nums2[j-1]`.
*   **State Transition:** If `nums1[i-1] == nums2[j-1]`, then `dp[i][j] = dp[i-1][j-1] + 1`. Otherwise, `dp[i][j] = 0`.
*   **Maximize:** Keep track of the maximum value in the DP table, which represents the maximum length of the common subarray.

*   **Time Complexity:** O(m\*n), where m and n are the lengths of nums1 and nums2 respectively.
*   **Space Complexity:** O(m\*n)

	
	# Code
	```python
	def findLength(nums1, nums2):
    """
    Finds the maximum length of a subarray that appears in both arrays.

    Args:
        nums1: The first integer array.
        nums2: The second integer array.

    Returns:
        The maximum length of a common subarray.
    """

    n = len(nums1)
    m = len(nums2)
    dp = [[0] * (m + 1) for _ in range(n + 1)]
    max_length = 0

    for i in range(1, n + 1):
        for j in range(1, m + 1):
            if nums1[i - 1] == nums2[j - 1]:
                dp[i][j] = dp[i - 1][j - 1] + 1
                max_length = max(max_length, dp[i][j])
            else:
                dp[i][j] = 0

    return max_length
	```