Add solution #1035

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,106 LeetCode solutions in JavaScript
+# 1,107 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -842,6 +842,7 @@
 1032|[Stream of Characters](./solutions/1032-stream-of-characters.js)|Hard|
 1033|[Moving Stones Until Consecutive](./solutions/1033-moving-stones-until-consecutive.js)|Medium|
 1034|[Coloring A Border](./solutions/1034-coloring-a-border.js)|Medium|
+1035|[Uncrossed Lines](./solutions/1035-uncrossed-lines.js)|Medium|
 1037|[Valid Boomerang](./solutions/1037-valid-boomerang.js)|Easy|
 1038|[Binary Search Tree to Greater Sum Tree](./solutions/1038-binary-search-tree-to-greater-sum-tree.js)|Medium|
 1041|[Robot Bounded In Circle](./solutions/1041-robot-bounded-in-circle.js)|Medium|

solutions/1035-uncrossed-lines.js

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+/**
+ * 1035. Uncrossed Lines
+ * https://leetcode.com/problems/uncrossed-lines/
+ * Difficulty: Medium
+ *
+ * You are given two integer arrays nums1 and nums2. We write the integers of nums1 and nums2
+ * (in the order they are given) on two separate horizontal lines.
+ *
+ * We may draw connecting lines: a straight line connecting two numbers nums1[i] and nums2[j]
+ * such that:
+ * - nums1[i] == nums2[j], and
+ * - the line we draw does not intersect any other connecting (non-horizontal) line.
+ *
+ * Note that a connecting line cannot intersect even at the endpoints (i.e., each number can only
+ * belong to one connecting line).
+ *
+ * Return the maximum number of connecting lines we can draw in this way.
+ */
+
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number}
+ */
+var maxUncrossedLines = function(nums1, nums2) {
+  const m = nums1.length;
+  const n = nums2.length;
+
+  const dp = new Array(m + 1).fill().map(() => new 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]) {
+        dp[i][j] = dp[i - 1][j - 1] + 1;
+      } else {
+        dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
+      }
+    }
+  }
+
+  return dp[m][n];
+};