<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:
        def isRepeated(n):
            s1, s2 = set(), set()
            for i in range(len(nums1)-n+1):
                s1.add(tuple(nums1[i:i+n]))

            for i in range(len(nums2)-n+1):
                s2.add(tuple(nums2[i:i+n]))
            return len(s1 & s2) > 0
        
        lo, hi = 0, max(len(nums1), len(nums2))
        while lo < hi:
            mid = (lo+hi+1) // 2
            if not isRepeated(mid):
                hi = mid-1
            else:
                lo = mid
        
        return lo
22300