<p>Given two integer arrays <code>nums1</code> and <code>nums2</code>, return <em>the maximum length of a subarray that appears in <strong>both</strong> arrays</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The repeated subarray with maximum length is [3,2,1].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The repeated subarray with maximum length is [0,0,0,0,0].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 100</code></li>
</ul>

class Solution:
    # If interst

    def LCStringTopDown(self, m: int, n: int, s1: List[int], s2: List[int], mem: List[List[Optional[int]]]) -> int:
        if m == 0 or n == 0:
            return 0

        if mem[m][n] is not None:
            return mem[m][n]

        if s1[m - 1] == s2[n - 1]:
            mem[m][n] = 1 + self.LCStringTopDown(m - 1, n - 1, s1, s2, mem)
            self.substrLenMax = max(self.substrLenMax, mem[m][n])
        else:
            mem[m][n] = 0
            
        
        self.LCStringTopDown(m, n - 1, s1, s2, mem)
        self.LCStringTopDown(m - 1, n, s1, s2, mem)
        return mem[m][n]

    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        self.substrLenMax = 0
        mem = [[None for _ in range(len(nums2) + 1)] for _ in range(len(nums1) + 1)]
        self.LCStringTopDown(len(nums1), len(nums2), nums1, nums2, mem)
        return self.substrLenMax

        
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