Introduction

In this blog post, we will explore how to solve the LeetCode problem "1377" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution to the frog jumping problem:

• High-Level Approach:
  • Use Depth-First Search (DFS) to explore the tree, keeping track of the probability and time taken to reach each node.
  • If the target is reached at the given time t, return the probability.
  • If the target is reached before t and the frog cannot move further, check if it stays at the target until time t.
• Complexity:
  • Runtime: O(N), where N is the number of vertices.
  • Storage: O(N) due to the adjacency list and recursion stack in the worst case.

Code

    def frogPosition(n, edges, t, target):
    """
    Calculates the probability of a frog being at a target vertex after a given time.

    Args:
        n: The number of vertices in the tree.
        edges: A list of edges representing the tree.
        t: The time after which to calculate the probability.
        target: The target vertex.

    Returns:
        The probability of the frog being at the target vertex after t seconds.
    """

    graph = [[] for _ in range(n + 1)]
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    probability = 0.0
    visited = [False] * (n + 1)

    def dfs(node, time, prob):
        nonlocal probability
        visited[node] = True

        if node == target:
            if time == t or (time < t and all(visited[neighbor] for neighbor in graph[node])):
                probability = prob
            return

        neighbors = [neighbor for neighbor in graph[node] if not visited[neighbor]]
        num_neighbors = len(neighbors)

        if num_neighbors == 0:
            return

        next_prob = prob / num_neighbors

        for neighbor in neighbors:
            dfs(neighbor, time + 1, next_prob)

    dfs(1, 0, 1.0)  # Start at vertex 1 with probability 1.0
    return probability