Uncrossed Lines
<p>You are given two integer arrays <code>nums1</code> and <code>nums2</code>. We write the integers of <code>nums1</code> and <code>nums2</code> (in the order they are given) on two separate horizontal lines.</p>

<p>We may draw connecting lines: a straight line connecting two numbers <code>nums1[i]</code> and <code>nums2[j]</code> such that:</p>

<ul>
	<li><code>nums1[i] == nums2[j]</code>, and</li>
	<li>the line we draw does not intersect any other connecting (non-horizontal) line.</li>
</ul>

<p>Note that a connecting line cannot intersect even at the endpoints (i.e., each number can only belong to one connecting line).</p>

<p>Return <em>the maximum number of connecting lines we can draw in this way</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1035.Uncrossed%20Lines/images/142.png" style="width: 400px; height: 286px;" />
<pre>
<strong>Input:</strong> nums1 = [1,4,2], nums2 = [1,2,4]
<strong>Output:</strong> 2
<strong>Explanation:</strong> We can draw 2 uncrossed lines as in the diagram.
We cannot draw 3 uncrossed lines, because the line from nums1[1] = 4 to nums2[2] = 4 will intersect the line from nums1[2]=2 to nums2[1]=2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [2,5,1,2,5], nums2 = [10,5,2,1,5,2]
<strong>Output:</strong> 3
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,3,7,1,7,5], nums2 = [1,9,2,5,1]
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 500</code></li>
	<li><code>1 &lt;= nums1[i], nums2[j] &lt;= 2000</code></li>
</ul>
Array; Dynamic Programming
/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var maxUncrossedLines = function (nums1, nums2) {
    const m = nums1.length;
    const n = nums2.length;
    const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(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;
            } else {
                f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
            }
        }
    }
    return f[m][n];
};
