Add solution #1466

Author: JoshCrozier

README.md

@@ -336,6 +336,7 @@
 1456|[Maximum Number of Vowels in a Substring of Given Length](./1456-maximum-number-of-vowels-in-a-substring-of-given-length.js)|Medium|
 1460|[Make Two Arrays Equal by Reversing Sub-arrays](./1460-make-two-arrays-equal-by-reversing-sub-arrays.js)|Easy|
 1464|[Maximum Product of Two Elements in an Array](./1464-maximum-product-of-two-elements-in-an-array.js)|Easy|
+1466|[Reorder Routes to Make All Paths Lead to the City Zero](./1466-reorder-routes-to-make-all-paths-lead-to-the-city-zero.js)|Medium|
 1470|[Shuffle the Array](./1470-shuffle-the-array.js)|Easy|
 1472|[Design Browser History](./1472-design-browser-history.js)|Medium|
 1475|[Final Prices With a Special Discount in a Shop](./1475-final-prices-with-a-special-discount-in-a-shop.js)|Easy|

solutions/1466-reorder-routes-to-make-all-paths-lead-to-the-city-zero.js

@@ -0,0 +1,49 @@
+/**
+ * 1466. Reorder Routes to Make All Paths Lead to the City Zero
+ * https://leetcode.com/problems/reorder-routes-to-make-all-paths-lead-to-the-city-zero/
+ * Difficulty: Medium
+ *
+ * There are n cities numbered from 0 to n - 1 and n - 1 roads such that there is only one way
+ * to travel between two different cities (this network form a tree). Last year, The ministry
+ * of transport decided to orient the roads in one direction because they are too narrow.
+ *
+ * Roads are represented by connections where connections[i] = [ai, bi] represents a road from
+ * city ai to city bi.
+ *
+ * This year, there will be a big event in the capital (city 0), and many people want to travel
+ * to this city.
+ *
+ * Your task consists of reorienting some roads such that each city can visit the city 0. Return
+ * the minimum number of edges changed.
+
+It's guaranteed that each city can reach city 0 after reorder.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} connections
+ * @return {number}
+ */
+var minReorder = function(n, connections) {
+  const graph = new Array(n).fill(null).map(() => []);
+
+  connections.forEach(([k, v]) => {
+    graph[k].push([v, 1]);
+    graph[v].push([k, 0]);
+  });
+
+  function dfs(node, parent) {
+    let count = 0;
+
+    graph[node].forEach(([k, v]) => {
+      if (k !== parent) {
+        count += v;
+        count += dfs(k, node);
+      }
+    });
+
+    return count;
+  }
+
+  return dfs(0, -1);
+};