<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:
        if len(nums1) < len(nums2):
            return self.findLength(nums2, nums1)
        
        m = len(nums1)
        n = len(nums2)
        currDP, prevDP = [0]*(n+1),  [0]*(n+1)
        ans = 0
        
        for i in range(m-1, -1, -1):
            for j in range(n-1, -1, -1):
                if nums1[i] == nums2[j]:
                    currDP[j] = prevDP[j+1] + 1
                    
                else:
                    currDP[j] = 0
                
                ans = max(ans, currDP[j])
                
            currDP, prevDP = prevDP, currDP
            
        return ans
    
16400