<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 check(self, value, nums1, nums2):
        if value == 0:
            return True
        strCurr = ""
        seen = set()
        for i in range(value):
            strCurr += str(nums1[i])
            strCurr += "-"
        seen.add(strCurr)
        for i in range(value, len(nums1)):
            dashPos = strCurr.index('-')
            strCurr = strCurr[dashPos+1:]
            strCurr += str(nums1[i])
            strCurr += "-"
            seen.add(strCurr)
        strCurr = ""
        for i in range(value):
            strCurr += str(nums2[i])
            strCurr += "-"
        if strCurr in seen:
            return True
        for i in range(value, len(nums2)):
            dashPos = strCurr.index('-')
            strCurr = strCurr[dashPos+1:]
            strCurr += str(nums2[i])
            strCurr += "-"

            if strCurr in seen:
                return True
        return False
    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        #binary search on the answer
        #10 10
        #101 0
        #1000 * 1000 * log(1000)
        lo = 0
        hi = min(len(nums1), len(nums2))
        while lo < hi:
            mid = (lo + hi + 1)//2
            if self.check(mid, nums1, nums2):
                lo = mid
            else:
                hi = mid-1
        return lo
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