Add solution #1436

Author: JoshCrozier

README.md

@@ -148,6 +148,7 @@
 1380|[Lucky Numbers in a Matrix](./1380-lucky-numbers-in-a-matrix.js)|Easy|
 1389|[Create Target Array in the Given Order](./1389-create-target-array-in-the-given-order.js)|Easy|
 1408|[String Matching in an Array](./1408-string-matching-in-an-array.js)|Easy|
+1436|[Destination City](./1436-destination-city.js)|Easy|
 1472|[Design Browser History](./1472-design-browser-history.js)|Medium|
 1598|[Crawler Log Folder](./1598-crawler-log-folder.js)|Easy|
 1880|[Check if Word Equals Summation of Two Words](./1880-check-if-word-equals-summation-of-two-words.js)|Easy|

solutions/1436-destination-city.js

@@ -0,0 +1,21 @@
+/**
+ * 1436. Destination City
+ * https://leetcode.com/problems/destination-city/
+ * Difficulty: Easy
+ *
+ * You are given the array paths, where paths[i] = [cityAi, cityBi] means there
+ * exists a direct path going from cityAi to cityBi. Return the destination city,
+ * that is, the city without any path outgoing to another city.
+ *
+ * It is guaranteed that the graph of paths forms a line without any loop, therefore,
+ * there will be exactly one destination city.
+ */
+
+/**
+ * @param {string[][]} paths
+ * @return {string}
+ */
+var destCity = function(paths) {
+  const set = new Set(paths.map(([a]) => a));
+  return paths.find(([_, b]) => !set.has(b))[1];
+};