<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>

static const int speedup = []{ios::sync_with_stdio(0); cin.tie(0); return 0;}();

class Solution {
public:
    int findLength(std::vector<int>& nums1, std::vector<int>& nums2) {
        std::string s1;
        s1.reserve(nums1.size());
        std::string s2;
        s2.reserve(nums2.size());
        for (int num : nums1)
            s1 += num + '0';
        for (int num : nums2)
            s2 += num + '0';
        auto range = [&s1, &s2](int size) {
            std::unordered_set<std::string_view> set;
            for (int i = 0; i <= s1.size() - size; ++i)
                set.emplace(std::string_view(s1.begin() + i, s1.begin() + i + size));
            for (int i = 0; i <= s2.size() - size; ++i) {
                if (set.contains(std::string_view(s2.begin() + i, s2.begin() + i + size)))
                    return true;
            }
            return false;
        };

        int max = std::min(nums1.size(), nums2.size()) + 1;
        int min = 0;
        while (min < max - 1) {
            int mid = min + (max - min) / 2;
            range(mid) ? min = mid : max = mid;
        }
        return min;
    }
};
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