Add solution #332

Author: JoshCrozier

0332-reconstruct-itinerary.js

@@ -0,0 +1,48 @@
+/**
+ * 332. Reconstruct Itinerary
+ * https://leetcode.com/problems/reconstruct-itinerary/
+ * Difficulty: Hard
+ *
+ * You are given a list of airline tickets where tickets[i] = [fromi, toi] represent the departure
+ * and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
+ *
+ * All of the tickets belong to a man who departs from "JFK", thus, the itinerary must begin with
+ * "JFK". If there are multiple valid itineraries, you should return the itinerary that has the
+ * smallest lexical order when read as a single string.
+ *
+ * For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].
+ *
+ * You may assume all tickets form at least one valid itinerary. You must use all the tickets once
+ * and only once.
+ */
+
+/**
+ * @param {string[][]} tickets
+ * @return {string[]}
+ */
+var findItinerary = function(tickets) {
+  const graph = new Map();
+  const itinerary = [];
+
+  for (const [from, to] of tickets) {
+    if (!graph.has(from)) {
+      graph.set(from, []);
+    }
+    graph.get(from).push(to);
+  }
+
+  for (const [_, destinations] of graph) {
+    destinations.sort().reverse();
+  }
+
+  dfs('JFK');
+
+  return itinerary.reverse();
+
+  function dfs(airport) {
+    while (graph.has(airport) && graph.get(airport).length) {
+      dfs(graph.get(airport).pop());
+    }
+    itinerary.push(airport);
+  }
+};

README.md

@@ -268,6 +268,7 @@
 329|[Longest Increasing Path in a Matrix](./0329-longest-increasing-path-in-a-matrix.js)|Hard|
 330|[Patching Array](./0330-patching-array.js)|Hard|
 331|[Verify Preorder Serialization of a Binary Tree](./0331-verify-preorder-serialization-of-a-binary-tree.js)|Medium|
+332|[Reconstruct Itinerary](./0332-reconstruct-itinerary.js)|Hard|
 334|[Increasing Triplet Subsequence](./0334-increasing-triplet-subsequence.js)|Medium|
 337|[House Robber III](./0337-house-robber-iii.js)|Medium|
 338|[Counting Bits](./0338-counting-bits.js)|Easy|