Leetcode #1377: Frog Position After T Seconds
In this guide, we solve Leetcode #1377 Frog Position After T Seconds in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an undirected tree consisting of n vertices numbered from 1 to n. A frog starts jumping from vertex 1.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Graph
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
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.
Python Solution
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
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.