<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>

const int MAX_N = 1000 + 10;
const int MAX_W = 1000 + 10;

int t[MAX_N + 1][MAX_W + 1];

class Solution {
public:
    int getRepeated(vector<int>& nums1, vector<int>& nums2, int& n, int& m) {
        int mx = INT_MIN;
        for (int i = 0; i < n + 1; i++) {
            for (int j = 0; j < m + 1; j++) {
                if (i == 0 or j == 0) {
                    t[i][j] = 0;
                }
                else if (nums1[i - 1] == nums2[j - 1]) {
                    t[i][j] = 1 + t[i - 1][j - 1];
                }
                else {
                    t[i][j] = 0;
                }
                mx = max(mx, t[i][j]);
            }
        }
        return mx;
    }

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

        return getRepeated(nums1, nums2, n, m);
    }
};
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