<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:
        #bsearch soln:
        #O(n log n)
        #Pick a fixed window size 
        #O(n) -- try all fixed windows of that size 
        #See if that subarray exists in bto
      


        def verify(size):
            arrs = set()
            for i in range(len(nums1) - size + 1):
                arrs.add(str(nums1[i:i+size]))
            for i in range(len(nums1) - size + 1):
                if str(nums2[i:i+size]) in arrs:
                    return True 
            return False 

        low = 0
        high = min(len(nums1), len(nums2)) 
        res = 0
        while low <= high:
            mid = (low + high) // 2 
            if verify(mid):
                low = mid + 1 
                res = max(res, mid)
            else:
                high = mid - 1 
        return res 

18200