Add solution #1266

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,394 LeetCode solutions in JavaScript
+# 1,395 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -966,6 +966,7 @@
 1261|[Find Elements in a Contaminated Binary Tree](./solutions/1261-find-elements-in-a-contaminated-binary-tree.js)|Medium|
 1262|[Greatest Sum Divisible by Three](./solutions/1262-greatest-sum-divisible-by-three.js)|Medium|
 1263|[Minimum Moves to Move a Box to Their Target Location](./solutions/1263-minimum-moves-to-move-a-box-to-their-target-location.js)|Hard|
+1266|[Minimum Time Visiting All Points](./solutions/1266-minimum-time-visiting-all-points.js)|Easy|
 1267|[Count Servers that Communicate](./solutions/1267-count-servers-that-communicate.js)|Medium|
 1268|[Search Suggestions System](./solutions/1268-search-suggestions-system.js)|Medium|
 1269|[Number of Ways to Stay in the Same Place After Some Steps](./solutions/1269-number-of-ways-to-stay-in-the-same-place-after-some-steps.js)|Hard|

solutions/1266-minimum-time-visiting-all-points.js

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+/**
+ * 1266. Minimum Time Visiting All Points
+ * https://leetcode.com/problems/minimum-time-visiting-all-points/
+ * Difficulty: Easy
+ *
+ * On a 2D plane, there are n points with integer coordinates points[i] = [xi, yi]. Return the
+ * minimum time in seconds to visit all the points in the order given by points.
+ *
+ * You can move according to these rules:
+ * - In 1 second, you can either:
+ *   - move vertically by one unit,
+ *   - move horizontally by one unit, or
+ *   - move diagonally sqrt(2) units (in other words, move one unit vertically then one unit
+ *     horizontally in 1 second).
+ * - You have to visit the points in the same order as they appear in the array.
+ * - You are allowed to pass through points that appear later in the order, but these do not
+ *   count as visits.
+ */
+
+/**
+ * @param {number[][]} points
+ * @return {number}
+ */
+var minTimeToVisitAllPoints = function(points) {
+  return points.reduce((steps, point, i) => steps + (!points[i + 1] ? 0 : Math.max(
+    Math.abs(point[0] - points[i + 1][0]), Math.abs(point[1] - points[i + 1][1])
+  )), 0);
+};