Add solution #1377

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,263 LeetCode solutions in JavaScript
+# 1,264 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1049,6 +1049,7 @@
 1374|[Generate a String With Characters That Have Odd Counts](./solutions/1374-generate-a-string-with-characters-that-have-odd-counts.js)|Easy|
 1375|[Number of Times Binary String Is Prefix-Aligned](./solutions/1375-number-of-times-binary-string-is-prefix-aligned.js)|Medium|
 1376|[Time Needed to Inform All Employees](./solutions/1376-time-needed-to-inform-all-employees.js)|Medium|
+1377|[Frog Position After T Seconds](./solutions/1377-frog-position-after-t-seconds.js)|Hard|
 1380|[Lucky Numbers in a Matrix](./solutions/1380-lucky-numbers-in-a-matrix.js)|Easy|
 1389|[Create Target Array in the Given Order](./solutions/1389-create-target-array-in-the-given-order.js)|Easy|
 1400|[Construct K Palindrome Strings](./solutions/1400-construct-k-palindrome-strings.js)|Medium|

solutions/1377-frog-position-after-t-seconds.js

@@ -0,0 +1,54 @@
+/**
+ * 1377. Frog Position After T Seconds
+ * https://leetcode.com/problems/frog-position-after-t-seconds/
+ * Difficulty: Hard
+ *
+ * Given an undirected tree consisting of n vertices numbered from 1 to n. A frog starts
+ * jumping from vertex 1. In one second, the frog jumps from its current vertex to another
+ * unvisited vertex if they are directly connected. The frog can not jump back to a visited
+ * vertex. In case the frog can jump to several vertices, it jumps randomly to one of them
+ * with the same probability. Otherwise, when the frog can not jump to any unvisited vertex,
+ * it jumps forever on the same 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.
+ *
+ * Return the probability that after t seconds the frog is on the vertex target. Answers
+ * within 10-5 of the actual answer will be accepted.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @param {number} t
+ * @param {number} target
+ * @return {number}
+ */
+var frogPosition = function(n, edges, t, target) {
+  const adjacencyList = Array.from({ length: n + 1 }, () => []);
+  for (const [a, b] of edges) {
+    adjacencyList[a].push(b);
+    adjacencyList[b].push(a);
+  }
+
+  return explore(1, 0, 1, new Set());
+
+  function explore(vertex, time, probability, visited) {
+    if (time > t) return 0;
+    if (vertex === target && time === t) return probability;
+    if (vertex === target && adjacencyList[vertex].every(v => visited.has(v))) return probability;
+
+    const unvisitedNeighbors = adjacencyList[vertex].filter(v => !visited.has(v));
+    if (unvisitedNeighbors.length === 0) return 0;
+
+    const nextProbability = probability / unvisitedNeighbors.length;
+    visited.add(vertex);
+
+    for (const neighbor of unvisitedNeighbors) {
+      const result = explore(neighbor, time + 1, nextProbability, visited);
+      if (result > 0) return result;
+    }
+
+    return 0;
+  }
+};