<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:
        nums1, nums2 = bytes(nums1), bytes(nums2)
        def has_subarray(m):
            sub_arr = {nums1[i:i+m+1] for i in range(len(nums1)-m)}
            return any(nums2[i:i+m+1] in sub_arr for i in range(len(nums2)-m))
        # return bisect_left(range(min(len(nums1), len(nums2))), 1, key = has_subarray)

        l, r = 0, min(len(nums1), len(nums2))
        while l < r:
            m = (l + r) // 2
            if has_subarray(m):
                l = m + 1
            else:
                r = m
        return l



        dp = [[0]*(len(nums2)+1) for _ in range(len(nums1)+1)]
        ans = 0
        for i, j in product(range(len(nums1)), range(len(nums2))):
            if nums1[i] == nums2[j]:
                dp[i+1][j+1] = dp[i][j] + 1
                ans = max(dp[i+1][j+1], ans)
        # print(dp)
        return ans

        
17400