Add solution #1443

Author: JoshCrozier

README.md

@@ -248,6 +248,7 @@
 1431|[Kids With the Greatest Number of Candies](./1431-kids-with-the-greatest-number-of-candies.js)|Easy|
 1436|[Destination City](./1436-destination-city.js)|Easy|
 1437|[Check If All 1's Are at Least Length K Places Away](./1437-check-if-all-1s-are-at-least-length-k-places-away.js)|Easy|
+1443|[Minimum Time to Collect All Apples in a Tree](./1443-minimum-time-to-collect-all-apples-in-a-tree.js)|Medium|
 1446|[Consecutive Characters](./1446-consecutive-characters.js)|Easy|
 1447|[Simplified Fractions](./1447-simplified-fractions.js)|Medium|
 1450|[Number of Students Doing Homework at a Given Time](./1450-number-of-students-doing-homework-at-a-given-time.js)|Easy|

solutions/1443-minimum-time-to-collect-all-apples-in-a-tree.js

@@ -0,0 +1,43 @@
+/**
+ * 1443. Minimum Time to Collect All Apples in a Tree
+ * https://leetcode.com/problems/minimum-time-to-collect-all-apples-in-a-tree/
+ * Difficulty: Medium
+ *
+ * Given an undirected tree consisting of n vertices numbered from 0 to n-1, which has some apples
+ * in their vertices. You spend 1 second to walk over one edge of the tree. Return the minimum time
+ * in seconds you have to spend to collect all apples in the tree, starting at vertex 0 and coming
+ * back to this vertex.
+ *
+ * The edges of the undirected tree are given in the array edges, where edges[i] = [ai, bi] means
+ * that exists an edge connecting the vertices ai and bi. Additionally, there is a boolean array
+ * hasApple, where hasApple[i] = true means that vertex i has an apple; otherwise, it does not
+ * have any apple.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @param {boolean[]} hasApple
+ * @return {number}
+ */
+var minTime = function(n, edges, hasApple) {
+  const map = new Map();
+  const seen = new Set();
+  let time = 0;
+
+  edges.forEach(([value, key]) => map.set(key, value));
+  for (let i = n - 1; i >= 0; i--) {
+    if (hasApple[i]) {
+      dfs(i);
+    }
+  }
+
+  function dfs(key) {
+    if (key === 0 || seen.has(key)) return;
+    seen.add(key);
+    time += 2;
+    dfs(map.get(key));
+  }
+
+  return time;
+};