#1377

Hard

auto_awesomeGraph traversal with depth-first search

LeetCode Problem Workspace

Frog Position After T Seconds

The problem asks for the probability that a frog reaches a target vertex after t seconds in a tree graph.

tree depth first search breadth first search graph

Problem Statement

You are given an undirected tree with n vertices, where each vertex is numbered from 1 to n. A frog begins its jump from vertex 1 and can jump to an unvisited vertex connected by an edge every second. If there are multiple vertices available to jump, the frog chooses one randomly, with equal probability. The frog cannot revisit any vertex, and if no unvisited vertices remain, it stays on its current vertex.

The problem asks you to compute the probability that, after t seconds, the frog is on the target vertex. The tree is defined by an array of edges, and your task is to return the probability of the frog being on the target vertex after t seconds. Answers within a 10^-5 error tolerance are acceptable.

Examples

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

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

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

1 <= n <= 100
edges.length == n - 1
edges[i].length == 2
1 <= ai, bi <= n
1 <= t <= 50
1 <= target <= n

Solution Approach

Depth-First Search (DFS) Approach

The primary approach to solving this problem is using DFS to traverse the tree. Starting from vertex 1, the frog has multiple options for its next jump, which can be recursively simulated. We maintain the probability at each vertex based on the frog's current position and the available vertices to jump to.

Recursive Probability Calculation

At each step, we calculate the probability of reaching the target by considering the frog's potential jumps. The frog can jump to multiple unvisited vertices, so the probability at each step is divided equally among the possible moves. This recursive approach calculates the probability at each vertex and time step until the desired time t is reached.

Memoization for Efficiency

To avoid recalculating probabilities for the same vertex and time combination, we use memoization to store the intermediate results. This significantly reduces the time complexity of the DFS approach and ensures that the solution runs efficiently within the problem's constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the final approach. The DFS approach with memoization has a time complexity of O(n * t), where n is the number of vertices, and t is the number of seconds. The space complexity is O(n * t) due to the storage of intermediate results in the memoization table.

What Interviewers Usually Probe

The candidate can efficiently solve the problem using DFS with memoization.
The candidate can explain how to handle the random jumps in the tree graph.
The candidate can discuss edge cases, such as when the frog cannot move or stays on a vertex.

Common Pitfalls or Variants

Common pitfalls

Forgetting to account for the case where the frog stays on the same vertex after all possible moves are visited.
Not properly dividing the probability when multiple vertices are available for the frog to jump to.
Failing to optimize the DFS approach with memoization, leading to excessive recomputation.

Follow-up variants

The frog jumps with a probability to any of the connected vertices, including cases where some vertices are visited earlier.
The tree can be directed, where the frog can only jump in one direction based on the tree structure.
Change the problem to ask for the probability of being on a vertex after a specific number of jumps, not time.

FAQ

What is the main algorithm to solve the 'Frog Position After T Seconds' problem?

The main algorithm involves depth-first search (DFS) to explore the tree and calculate the probability of the frog being on the target vertex after t seconds, with memoization to optimize performance.

How can I optimize my solution for the 'Frog Position After T Seconds' problem?

You can optimize your solution by using memoization to store intermediate results and avoid redundant calculations. This reduces the time complexity of the DFS traversal.

What is the time complexity of the solution for the 'Frog Position After T Seconds' problem?

The time complexity of the DFS approach with memoization is O(n * t), where n is the number of vertices and t is the number of seconds.

Can the 'Frog Position After T Seconds' problem be solved without using DFS?

While DFS is the most natural approach for this problem, other techniques such as breadth-first search (BFS) can also be used, but DFS is typically more efficient for tree traversal.

What are common mistakes to avoid when solving the 'Frog Position After T Seconds' problem?

Common mistakes include not properly handling the case where the frog stays on a vertex, failing to divide the probability when there are multiple possible jumps, and not using memoization to optimize the DFS approach.

terminal

Solution

Solution 1: BFS

First, based on the undirected tree edges given in the problem, we construct an adjacency list $g$, where $g[u]$ represents all adjacent vertices of vertex $u$.

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class Solution:
    def frogPosition(
        self, n: int, edges: List[List[int]], t: int, target: int
    ) -> float:
        g = defaultdict(list)
        for u, v in edges:
            g[u].append(v)
            g[v].append(u)
        q = deque([(1, 1.0)])
        vis = [False] * (n + 1)
        vis[1] = True
        while q and t >= 0:
            for _ in range(len(q)):
                u, p = q.popleft()
                cnt = len(g[u]) - int(u != 1)
                if u == target:
                    return p if cnt * t == 0 else 0
                for v in g[u]:
                    if not vis[v]:
                        vis[v] = True
                        q.append((v, p / cnt))
            t -= 1
        return 0

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