Introduction

In this blog post, we will explore how to solve the LeetCode problem "2467" 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

There is an undirected tree with n nodes labeled from 0 to n - 1, rooted at node 0. You are given a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. At every node i, there is a gate. You are also given an array of even integers amount, where amount[i] represents: the price needed to open the gate at node i, if amount[i] is negative, or, the cash reward obtained on opening the gate at node i, otherwise. The game goes on as follows: Initially, Alice is at node 0 and Bob is at node bob. At every second, Alice and Bob each move to an adjacent node. Alice moves towards some leaf node, while Bob moves towards node 0. For every node along their path, Alice and Bob either spend money to open the gate at that node, or accept the reward. Note that: If the gate is already open, no price will be required, nor will there be any cash reward. If Alice and Bob reach the node simultaneously, they share the price/reward for opening the gate there. In other words, if the price to open the gate is c, then both Alice and Bob pay c / 2 each. Similarly, if the reward at the gate is c, both of them receive c / 2 each. If Alice reaches a leaf node, she stops moving. Similarly, if Bob reaches node 0, he stops moving. Note that these events are independent of each other. Return the maximum net income Alice can have if she travels towards the optimal leaf node. Example 1: Input: edges = [[0,1],[1,2],[1,3],[3,4]], bob = 3, amount = [-2,4,2,-4,6] Output: 6 Explanation: The above diagram represents the given tree. The game goes as follows: - Alice is initially on node 0, Bob on node 3. They open the gates of their respective nodes. Alice's net income is now -2. - Both Alice and Bob move to node 1. Since they reach here simultaneously, they open the gate together and share the reward. Alice's net income becomes -2 + (4 / 2) = 0. - Alice moves on to node 3. Since Bob already opened its gate, Alice's income remains unchanged. Bob moves on to node 0, and stops moving. - Alice moves on to node 4 and opens the gate there. Her net income becomes 0 + 6 = 6. Now, neither Alice nor Bob can make any further moves, and the game ends. It is not possible for Alice to get a higher net income. Example 2: Input: edges = [[0,1]], bob = 1, amount = [-7280,2350] Output: -7280 Explanation: Alice follows the path 0->1 whereas Bob follows the path 1->0. Thus, Alice opens the gate at node 0 only. Hence, her net income is -7280. Constraints: 2 <= n <= 105 edges.length == n - 1 edges[i].length == 2 0 <= ai, bi < n ai != bi edges represents a valid tree. 1 <= bob < n amount.length == n amount[i] is an even integer in the range [-104, 104].

Explanation

Here's the breakdown of the approach and the Python code:

• High-Level Approach:

  • Find the path of Bob from his initial node to the root (node 0). Store the time/distance of each node from Bob's starting point to the root.
  • Perform a Depth-First Search (DFS) from the root (node 0) for Alice. During the DFS, track Alice's path and the net income she accumulates.
  • For each node Alice visits, check if Bob also visits it. If they meet at the same time, divide the amount by 2. If Bob has already visited the node, Alice gets nothing. If Alice reaches before Bob, Alice gets the full amount.

• Complexity:

  • Runtime Complexity: O(N), where N is the number of nodes.
  • Storage Complexity: O(N)

Code

    from collections import defaultdict

def max_alice_income(edges, bob, amount):
    """
    Calculates the maximum net income Alice can have in the game.

    Args:
      edges: A 2D integer array representing the tree edges.
      bob: The initial node of Bob.
      amount: An array of even integers representing the gate amounts.

    Returns:
      The maximum net income Alice can have.
    """

    n = len(amount)
    graph = defaultdict(list)
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    bob_path_time = [-1] * n  # -1 means unvisited. Stores time from bob to that node
    bob_path_time[bob] = 0
    bob_path = []

    def dfs_bob(node, parent):
        bob_path.append(node)
        if node == 0:
            return True
        for neighbor in graph[node]:
            if neighbor != parent:
                bob_path_time[neighbor] = bob_path_time[node] + 1
                if dfs_bob(neighbor, node):
                    return True
        bob_path.pop() # backtrack if node not on bob's path
        return False

    dfs_bob(bob, -1)


    max_income = float('-inf')

    def dfs_alice(node, parent, current_income, depth):
        nonlocal max_income

        if depth == 0: #Alice starts at node 0
            if bob_path_time[node] == 0:
                current_income += amount[node] // 2
            elif bob_path_time[node] > 0:
                current_income += amount[node]
            else: # bob never reaches this node.
                current_income += amount[node]

        else:
          if bob_path_time[node] == depth:
              current_income += amount[node] // 2
          elif bob_path_time[node] > depth or bob_path_time[node] == -1: #bob arrives later or not at all
              current_income += amount[node]
          #else: bob reached it earlier, so do nothing.

        is_leaf = True
        for neighbor in graph[node]:
            if neighbor != parent:
                is_leaf = False
                dfs_alice(neighbor, node, current_income, depth + 1)

        if is_leaf:
            max_income = max(max_income, current_income)

    dfs_alice(0, -1, 0, 0) # start node, prev node, initial income, initial depth

    return max_income