How do you solve Frog Position After T Seconds on LeetCode?

Frog Position After T Seconds (LeetCode #1377) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The frog's movement is probabilistic based on available unvisited vertices.

What is the brute force solution for Frog Position After T Seconds?

The brute force approach for Frog Position After T Seconds is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating every possible path the frog can take for 't' seconds. We calculate the probability of reaching the target vertex by exploring all paths recursively. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Frog Position After T Seconds?

Frog Position After T Seconds uses the following concepts: Tree, Depth-First Search, Breadth-First Search, Graph Theory. The recommended patterns to study are: Depth-First Search, Graph Traversal.

What are the interview tips for Frog Position After T Seconds?

Clarify the problem constraints and edge cases before coding. Think about how to represent the tree structure effectively. Practice explaining your thought process as you code.

What are common mistakes when solving Frog Position After T Seconds?

Not accounting for the case when the frog has no unvisited neighbors. Forgetting to divide the probability by the number of unvisited neighbors.

#1377

Frog Position After T Seconds

Hard

TreeDepth-First SearchBreadth-First SearchGraph TheoryDepth-First SearchGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a depth-first search (DFS) approach with memoization to avoid recalculating probabilities for the same state. This significantly reduces redundant calculations and improves efficiency.

⚙️

Algorithm

4 steps

1Step 1: Build the graph from the edges.
2Step 2: Use DFS to explore paths from vertex 1, keeping track of time left and current probability.
3Step 3: If the time is zero, check if the current vertex is the target; if yes, return the probability.
4Step 4: For each unvisited neighbor, recursively call DFS, dividing the probability by the number of unvisited neighbors.

solution.py22 lines

1# Full working Python code
2from collections import defaultdict
3
4
5def frogPosition(n, edges, t, target):
6    graph = defaultdict(list)
7    for u, v in edges:
8        graph[u].append(v)
9        graph[v].append(u)
10
11    def dfs(vertex, time_left, prob, visited):
12        if time_left == 0:
13            return prob if vertex == target else 0
14        visited.add(vertex)
15        total_prob = 0
16        unvisited_neighbors = [v for v in graph[vertex] if v not in visited]
17        for neighbor in unvisited_neighbors:
18            total_prob += dfs(neighbor, time_left - 1, prob / len(unvisited_neighbors), visited)
19        visited.remove(vertex)
20        return total_prob
21
22    return dfs(1, t, 1, set())

ℹ

Complexity note: The optimal solution runs in O(n) time because we effectively traverse each vertex once, and the space complexity is O(n) due to the storage of the graph and visited set.

1The frog's movement is probabilistic based on available unvisited vertices.
2The tree structure ensures that each vertex has a unique path, simplifying traversal.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.