Introduction

In this blog post, we will explore how to solve the LeetCode problem "1443" 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 0 to n-1, which has some apples in their vertices. You spend 1 second to walk over one edge of the tree. Return the minimum time in seconds you have to spend to collect all apples in the tree, starting at vertex 0 and coming back to this 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. Additionally, there is a boolean array hasApple, where hasApple[i] = true means that vertex i has an apple; otherwise, it does not have any apple. Example 1: Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,true,true,false] Output: 8 Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows. Example 2: Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,false,true,false] Output: 6 Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows. Example 3: Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,false,false,false,false,false] Output: 0 Constraints: 1 <= n <= 105 edges.length == n - 1 edges[i].length == 2 0 <= ai < bi <= n - 1 hasApple.length == n

Explanation

Here's the breakdown:

• High-Level Approach:

  • Build an adjacency list representation of the tree.
  • Perform a Depth-First Search (DFS) to traverse the tree. The DFS returns the cost (time) to collect apples in the subtree rooted at each node.
  • The cost for a subtree is calculated only if the subtree contains an apple (either directly at the node or in its children). We also avoid double-counting edges by only including the cost of going to a child and back if that child's subtree has apples.

• Complexity:

  • Runtime: O(N), where N is the number of nodes in the tree.
  • Storage: O(N) due to the adjacency list and recursion stack.

Code

    def minTime(n: int, edges: list[list[int]], hasApple: list[bool]) -> int:
    """
    Calculates the minimum time to collect all apples in a tree, starting and returning to vertex 0.

    Args:
        n: The number of vertices in the tree.
        edges: A list of edges representing the tree.
        hasApple: A boolean array indicating whether each vertex has an apple.

    Returns:
        The minimum time to collect all apples.
    """

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

    def dfs(node: int, parent: int) -> int:
        """
        Performs a Depth-First Search to calculate the cost of collecting apples in the subtree rooted at 'node'.

        Args:
            node: The current node being visited.
            parent: The parent of the current node.

        Returns:
            The cost (time) to collect apples in the subtree rooted at 'node'.
        """

        cost = 0
        for neighbor in adj[node]:
            if neighbor != parent:
                child_cost = dfs(neighbor, node)
                if child_cost > 0 or hasApple[neighbor]:
                    cost += 2 + child_cost  # 2 for going to neighbor and back

        return cost

    return dfs(0, -1)