<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>& a, vector<int>& b) {
        vector<vector<int>> dp(a.size()+1,vector<int>(b.size()+1,-1)) ;
        int ans=INT_MIN;
        for(int i=0;i<=a.size();i++){
            for(int j=0;j<=b.size();j++){
                if(i==0||j==0)
                dp[i][j]=0;
            }
        }

         for(int i=1;i<=a.size();i++){
            for(int j=1;j<=b.size();j++){
                if(a[i-1]==b[j-1])
                dp[i][j]=1+dp[i-1][j-1];
                else
                dp[i][j]=0;
            }
        }

          for(int i=1;i<=a.size();i++){
            for(int j=1;j<=b.size();j++){
               ans=max(dp[i][j],ans);
            }
        }

        return ans;
    }
};
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