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

                }
                ans = max(ans, dp[i][j]);
            }
        }
        return ans;
    }
};
50379