Introduction

In this blog post, we will explore how to solve the LeetCode problem "2973" 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

You are given an undirected tree with n nodes labeled from 0 to n - 1, and 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. You are also given a 0-indexed integer array cost of length n, where cost[i] is the cost assigned to the ith node. You need to place some coins on every node of the tree. The number of coins to be placed at node i can be calculated as: If size of the subtree of node i is less than 3, place 1 coin. Otherwise, place an amount of coins equal to the maximum product of cost values assigned to 3 distinct nodes in the subtree of node i. If this product is negative, place 0 coins. Return an array coin of size n such that coin[i] is the number of coins placed at node i. Example 1: Input: edges = [[0,1],[0,2],[0,3],[0,4],[0,5]], cost = [1,2,3,4,5,6] Output: [120,1,1,1,1,1] Explanation: For node 0 place 6 5 4 = 120 coins. All other nodes are leaves with subtree of size 1, place 1 coin on each of them. Example 2: Input: edges = [[0,1],[0,2],[1,3],[1,4],[1,5],[2,6],[2,7],[2,8]], cost = [1,4,2,3,5,7,8,-4,2] Output: [280,140,32,1,1,1,1,1,1] Explanation: The coins placed on each node are: - Place 8 7 5 = 280 coins on node 0. - Place 7 5 4 = 140 coins on node 1. - Place 8 2 2 = 32 coins on node 2. - All other nodes are leaves with subtree of size 1, place 1 coin on each of them. Example 3: Input: edges = [[0,1],[0,2]], cost = [1,2,-2] Output: [0,1,1] Explanation: Node 1 and 2 are leaves with subtree of size 1, place 1 coin on each of them. For node 0 the only possible product of cost is 2 1 -2 = -4. Hence place 0 coins on node 0. Constraints: 2 <= n <= 2 * 104 edges.length == n - 1 edges[i].length == 2 0 <= ai, bi < n cost.length == n 1 <= |cost[i]| <= 104 The input is generated such that edges represents a valid tree.

Explanation

Here's a breakdown of the approach, complexity analysis, and the Python code:

• High-Level Approach:

  • Perform Depth-First Search (DFS) to traverse the tree.
  • At each node, compute the subtree size and maintain a sorted list of the costs of nodes within that subtree (up to a certain limit, like 5).
  • Calculate the maximum product of 3 distinct costs within the subtree and determine the coin value for the node.

• Complexity:

  • Runtime: O(n log k), where n is the number of nodes and k is the maximum size of the sorted list of costs we maintain at each node (k = 5 in this solution). The log k factor comes from sorting or maintaining a sorted list.
  • Storage: O(n), for storing the adjacency list representation of the tree and the coin values for each node.

Code

    def place_coins(edges, cost):
    n = len(cost)
    graph = [[] for _ in range(n)]
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    coin = [0] * n
    subtree_costs = [[] for _ in range(n)]  # Store top 5 costs (pos/neg) in subtree

    def dfs(node, parent):
        subtree_costs[node].append(cost[node])
        subtree_size = 1

        for neighbor in graph[node]:
            if neighbor != parent:
                subtree_size += dfs(neighbor, node)
                subtree_costs[node].extend(subtree_costs[neighbor])

        # Keep only top 5 largest and 2 smallest values in subtree costs
        subtree_costs[node].sort(reverse=True)
        subtree_costs[node] = subtree_costs[node][:5]
        negatives = sorted([x for x in subtree_costs[node] if x < 0])
        subtree_costs[node].extend(negatives[:2])

        if subtree_size < 3:
            coin[node] = 1
        else:
            max_product = float('-inf')
            nums = list(set(subtree_costs[node])) #Remove duplicates

            if len(nums) >= 3:
                for i in range(len(nums)):
                    for j in range(i + 1, len(nums)):
                        for k in range(j + 1, len(nums)):
                            product = nums[i] * nums[j] * nums[k]
                            max_product = max(max_product, product)

                coin[node] = max(0, max_product)

        return subtree_size

    dfs(0, -1)
    return coin