<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 m= nums1.size(),n=nums2.size();
        vector<vector<int>>dp(m+1,vector<int>(n+1,0));
        int ans =0;
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(nums1[i-1]==nums2[j-1]) dp[i][j] = 1+dp[i-1][j-1];
                ans = max(ans,dp[i][j]);
            }
        }
        return ans;
    }
};
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