Is Frog Position After T Seconds easy or hard?

The problem is labeled Hard because it combines tree traversal with probability reasoning and time constraints. Many candidates implement DFS or BFS correctly but miss edge cases such as staying at a leaf when time remains. Handling these conditions correctly is what makes the problem challenging.

Frog Position After T Seconds Python/Java solution

Python and Java implementations usually rely on recursive DFS that carries three parameters: current node, elapsed time, and probability. JavaScript and C# implementations often use BFS with a queue storing node and probability pairs. Both styles achieve O(n) time complexity.

How to solve Frog Position After T Seconds in O(n)?

Build an adjacency list for the tree and simulate the frog’s movement using DFS or BFS. At each node, count unvisited neighbors and divide the current probability equally among them. Stop when time reaches t or when a leaf node is reached. Since each node is processed once, the algorithm runs in O(n).

What is the best approach for Frog Position After T Seconds?

The most common solution uses DFS with probability propagation on a tree. Each node distributes its probability equally among unvisited neighbors while tracking elapsed time. This approach runs in O(n) time because every node and edge is processed once. BFS with level tracking provides an equally optimal alternative.

Is Frog Position After T Seconds asked at Google/Amazon/Meta?

Tree traversal and probability propagation problems frequently appear in interviews at companies like Google, Amazon, and Meta. While this exact problem may not always appear verbatim, variations involving BFS or DFS on trees with state tracking are common interview patterns.

What data structure is used in Frog Position After T Seconds?

The core data structure is an adjacency list representing the tree. The traversal uses either a recursion stack for DFS or a queue for BFS. A visited set or parent tracking ensures the frog never revisits nodes, preserving the tree traversal constraint.

What is the time complexity of Frog Position After T Seconds?

The optimal time complexity is O(n), where n is the number of nodes in the tree. Both DFS and BFS traverse each node and edge at most once while distributing probabilities to neighbors. Space complexity is also O(n) due to adjacency lists and recursion stack or queue storage.

Description Hints 2 Solution Video 21:39

1377. Frog Position After T Seconds

Hard38.8% acceptance2 companies

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] = [a_i, b_i] means that exists an edge connecting the vertices a_i and b_i.

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.

Example 1:

Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 2, target = 4
Output: 0.16666666666666666 
Explanation: The figure above shows the given graph. The frog starts at vertex 1, jumping with 1/3 probability to the vertex 2 after second 1 and then jumping with 1/2 probability to vertex 4 after second 2. Thus the probability for the frog is on the vertex 4 after 2 seconds is 1/3 * 1/2 = 1/6 = 0.16666666666666666.

Example 2:

Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 1, target = 7
Output: 0.3333333333333333
Explanation: The figure above shows the given graph. The frog starts at vertex 1, jumping with 1/3 = 0.3333333333333333 probability to the vertex 7 after second 1.

Constraints: