<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 findLength(vector<int>& nums1, vector<int>& nums2) {
        int n = nums1.size();
        int m = nums2.size();

        vector<vector<int>> dp(n,vector<int>(m,-1));
        helper(nums1, nums2, 0, 0, dp);

        int ans = 0;
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                ans = max(ans, dp[i][j]);
            }
        }

        return ans;
    }
    
    //helper i,j will return the length of the longest subarray starting at i in n1 and j in n2
    int helper(vector<int>& n1, vector<int>& n2, int i, int j,auto& dp){
        if(i==n1.size() || j==n2.size())return 0;

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

        int ans = 0;
        if(n1[i]==n2[j]){
            ans = 1 + helper(n1, n2, i+1, j+1, dp);
        }

        helper(n1, n2, i, j+1, dp);
        helper(n1, n2, i+1, j, dp);

        return dp[i][j]=ans; 
    }
};
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