<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 {
public:
    string s1,s2; int n1,n2;
    bool can(int k){
        if(k==0)return true;
        unordered_set<string>st;
        for(int i=0;i+k<=n2;i++){
            st.insert(s2.substr(i,k));
        }
        for(int i=0;i+k<=n1;i++){
            string s=s1.substr(i,k);
            if(st.find(s)!=st.end())
                return true;
        }
        return false;
    }
    int findLength(vector<int>& nums1, vector<int>& nums2) {
        for(auto x:nums1)
            s1.push_back(x+'0');
        for(auto x:nums2)
            s2.push_back(x+'0');
        n1=s1.size(); n2=s2.size();
        int l=0,r=min(s1.size(),s2.size())+1;
        while(l<r){
            int mid=(l+r)/2;
            if(can(mid))
                l=mid+1;
            else
                r=mid;
        }
        return l-1;
    }
};
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