<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:
    int dp[10001][1001];
    int ans=0;
    int rec(int i, int j, int m , int n, vector<int>& nums1, vector<int>& nums2){
        if(i>=m || j>=n) return 0;

        if(dp[i][j]!=-1) return dp[i][j];

        int p=0;
        if(nums1[i]==nums2[j]){
            p=1+rec(i+1,j+1,m,n,nums1,nums2);
        }
        ans=max(ans,p);
        rec(i,j+1,m,n,nums1,nums2);
        rec(i+1,j,m,n,nums1,nums2);

        return dp[i][j]=p;
    }
    int findLength(vector<int>& nums1, vector<int>& nums2) {
        int m=nums1.size();
        int n=nums2.size();
        memset(dp,-1,sizeof(dp));
        rec(0,0,m,n,nums1,nums2);
        return ans;
    }
};
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