<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:
    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        dp=[[0 for _ in range(len(nums1))] for _ in range(len(nums2))]
        for j in range(len(nums2)):
            for i in range(len(nums1)):
                if nums1[i]==nums2[j]:
                    prev=dp[j-1][i-1] if (i>0 and j>0) else 0
                    dp[j][i]=max(dp[j][i],1+prev)
        return max([max(i) for i in dp])

        # dp={}
        # def dfs(i,j):
        #     if (i,j) in dp:
        #         return dp[(i,j)]
        #     if i>=len(nums1) or j>=len(nums2):
        #         return 0
        #     res=0
        #     if nums1[i]==nums2[j]:
        #         res=max(res, 1+dfs(i+1,j+1))
        #     dfs(i+1,j)
        #     dfs(i,j+1)
        #     dp[(i,j)]=res
        #     return res
        # dfs(0,0)
        # return max(dp.values())


        
42100