<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>

int dp[1001][1001];
class Solution {
public:
    int LCS(vector<int>& x, vector<int>& y, int n, int m) {

        int maxi = 0;
        // Initialize the dp array to 0
        for (int i = 0; i <= n; i++) {
            dp[i][0] = 0;
        }
        for (int j = 0; j <= m; j++) {
            dp[0][j] = 0;
        }

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (x[i - 1] == y[j - 1]) {
                    dp[i][j] = 1 + dp[i - 1][j - 1];
                    maxi = max(maxi, dp[i][j]);
                } else {
                    dp[i][j] = 0;
                }
            }
        }
        return maxi;
    }

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