<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, nums2):
        dp = [[0] * (len(nums1)+1) for i in range((len(nums2)+1))]
        for i in range(len(nums2)-1,-1,-1):
            for j in range(len(nums1)-1,-1,-1):
                if nums2[i] == nums1[j]:
                    dp[i][j] = 1
        for i in range(len(nums2)-1,-1,-1):
            for j in range(len(nums1)-1,-1,-1):
                if nums2[i] == nums1[j]:
                    dp[i][j] = dp[i][j] + dp[i+1][j+1]
        #         # else:
        #         #     dp[i][j] = dp[i+1][j+1]
        return max(map(max,dp))

41000