<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, A: List[int], B: List[int]) -> int:
        M = (1 << 64) - 1
        base = 101

        def rolling_hash(arr, x):
            Bx = pow(base, x, M)
            ans = set()
            cur = 0
            for i in range(x):
                cur = (cur * base + arr[i]) % M
            ans.add(cur)
            for i in range(x, len(arr)):
                cur = (cur * base - arr[i - x] * Bx + arr[i]) % M
                ans.add(cur)
            return ans

        # find min x that no repeated subarray of size x in both array
        def condition(x):
            A_hash = rolling_hash(A, x)
            B_hash = rolling_hash(B, x)
            if A_hash & B_hash:
                return False
            return True

        m, n = len(A), len(B)
        l, r = 1, min(m, n) + 1
        while l < r:
            mid = (l + r) // 2
            if condition(mid):
                r = mid
            else:
                l = mid + 1
        return l - 1


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