<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>

from functools import cache
from bisect import bisect_left, bisect

class Solution:
    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        nums1 = tuple(nums1)
        nums2 = tuple(nums2)

        def appears_in(subarray: tuple[int, ...], array: tuple[int, ...]) -> bool:
            return any(
                subarray == array[i : i + len(subarray)]
                for i in range(len(array))
            )

        def subarrays(array: tuple[int], length: int) -> set[tuple[int]]:
            return {
                tuple(array[i : i + length])
                for i in range(len(array) - length + 1)
            }

        return len(nums1) - bisect_left(
            range(len(nums1), 0, -1),
            True,
            key=lambda length: bool(subarrays(nums1, length) & subarrays(nums2, length))
        )
        return max(
            (
                len(subarray)
                for length in range(1, len(nums1) + 1)
                for subarray in subarrays(nums1, length) & subarrays(nums2, length)
            ),
            default=0,
        )
22500