<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>

bool check(int k, string &s1, string &s2){
    set<string>s;
    for(int i = 0; i <= s1.size()-k; ++i){
        s.insert(s1.substr(i,k));
    }
    for(int i = 0; i <= s2.size()-k; ++i){
        if( s.find(s2.substr(i,k)) != s.end() ) return true;
    }
    return false;
}


class Solution {
public:
    int findLength(vector<int>& n1 , vector<int>& n2) {
        string s1,s2;
        for(int i = 0; i < n1.size(); ++i){
            s1.push_back(n1[i] + '0');
        }
        for(int i = 0; i < n2.size(); ++i){
            s2.push_back(n2[i] + '0');
        }
        int l = -1, r = min(n1.size(),n2.size()) + 1;

        while(r - l > 1){
            int mid = (l+r)/2;
            if(check(mid, s1, s2)) l = mid;
            else r = mid;
        }

        return l;
    }
};
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