Maximum Length of Repeated Subarray
<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>
Array; Binary Search; Dynamic Programming; Sliding Window; Hash Function; Rolling Hash
/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findLength = function (nums1, nums2) {
    const m = nums1.length;
    const n = nums2.length;
    const f = Array.from({ length: m + 1 }, _ => new Array(n + 1).fill(0));
    let ans = 0;
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            if (nums1[i - 1] == nums2[j - 1]) {
                f[i][j] = f[i - 1][j - 1] + 1;
                ans = Math.max(ans, f[i][j]);
            }
        }
    }
    return ans;
};
