<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:
        ROWS = len(nums1)
        COLS = len(nums2)
        dp = [[0] * COLS for r in range(ROWS)]

        for i in range(ROWS):
            if nums2[COLS - 1] == nums1[i]:
                dp[i][COLS - 1] = 1
        
        for j in range(COLS):
            if nums1[ROWS - 1] == nums2[j]:
                dp[ROWS - 1][j] = 1
        max_amount = -math.inf
        for i in range(ROWS - 2, -1, -1):
            for j in range(COLS - 2, -1, -1):
                if nums1[i] == nums2[j]:
                    dp[i][j] = 1 + dp[i + 1][j + 1]
                # else:
                #     dp[i][j] = dp[i + 1][j + 1]
        
        for i in range(ROWS):
            for j in range(COLS):
                max_amount = max(max_amount, dp[i][j])
        return max_amount

        
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