<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 check(k):
            if k == 0:
                return True
            d = set()
            s = []
            for i in range(k):
                s+=[nums2[i]]
            d.add(tuple(s))
            for i in range(len(nums2)-k):
                s=s[1:]+[nums2[i+k]]
                d.add(tuple(s))
            s = []
            for i in range(k):
                s+=[nums1[i]]
            if tuple(s) in d:
                return True
            for i in range(len(nums1)-k):
                s=s[1:]+[nums1[i+k]]
                if tuple(s) in d:
                    return True
            return False
        
        l = 0
        r = min(len(nums1), len(nums2))+1
        p = [None]*(r)
        while l<r:
            mid = (l+r)//2
            p[mid] = check(mid)
            if check(mid):
                l = mid+1
            else:
                r = mid
        print(l,r)
        print(p)
        return (l+r)//2-1
20100