<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 dpp[1001][1001];
    int subarr( int i , int j, vector<int> &nums1 , vector<int>&nums2){
        if( i >= nums1.size() || j >= nums2.size() ) return 0;
        if( nums1[i] != nums2[j] ) return 0;
        if( dpp[i][j] != -1 ) return dpp[i][j];
        return dpp[i][j] = 1 + subarr(i+1,j+1,nums1,nums2);
    }
    int dp[1001][1001];
    int maxlen( int i , int j , vector<int> &nums1 , vector<int>&nums2 ){
        if( i == nums1.size() ) return 0;
        if( j == nums2.size() ) return 0;
        if( dp[i][j] != -1 ) return dp[i][j];
        int ans = 0;
        if( nums1[i] == nums2[j] ) {
            ans = subarr(i,j,nums1,nums2);
        }
        return dp[i][j] = max( { ans ,maxlen(i+1,j,nums1,nums2) , maxlen(i,j+1,nums1,nums2)} );
    }
    int findLength(vector<int>& nums1, vector<int>& nums2) {
        memset(dp,-1,sizeof(dp));
        memset(dpp,-1,sizeof(dpp));
        return maxlen(0,0,nums1,nums2);
    }
};
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