<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:
        # Not optimal
        dp = [[-1] * (len(nums1) + 1) for _ in range(len(nums2) + 1)]
        curMax = 0
        for i in range(len(nums1)):
            if i + curMax > len(nums1):
                break
            for j in range(len(nums2)):
                if dp[i][j] != -1:
                    continue
                if j + curMax > len(nums2):
                    break
                temp = 0
                while i+temp < len(nums1) and j+temp < len(nums2) and nums1[i+temp] == nums2[j+temp]:
                    temp += 1
                dp[i][j] = temp
                if temp > curMax:
                    curMax = temp
                j = j + temp - 1
        return curMax

27700