<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 valid(x):
            set1, set2 = set(), set()

            for i in range(len(nums1) - x + 1):
                set1.add(tuple(nums1[i:i + x]))

            for i in range(len(nums2) - x + 1):
                set2.add(tuple(nums2[i:i + x]))

            for sub1 in set1:
                for sub2 in set2:
                    if sub1 == sub2:
                        return True

            return False


        s = 0
        e = min(len(nums1), len(nums2))
        res = 0

        while s <= e:
            mid = (s + e) // 2

            if valid(mid):
                res = mid
                s = mid + 1
            else:
                e = mid - 1


        return res




                
            


22000