Introduction

In this blog post, we will explore how to solve the LeetCode problem "2925" 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, 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 values of length n, where values[i] is the value associated with the ith node. You start with a score of 0. In one operation, you can: Pick any node i. Add values[i] to your score. Set values[i] to 0. A tree is healthy if the sum of values on the path from the root to any leaf node is different than zero. Return the maximum score you can obtain after performing these operations on the tree any number of times so that it remains healthy. Example 1: Input: edges = [[0,1],[0,2],[0,3],[2,4],[4,5]], values = [5,2,5,2,1,1] Output: 11 Explanation: We can choose nodes 1, 2, 3, 4, and 5. The value of the root is non-zero. Hence, the sum of values on the path from the root to any leaf is different than zero. Therefore, the tree is healthy and the score is values[1] + values[2] + values[3] + values[4] + values[5] = 11. It can be shown that 11 is the maximum score obtainable after any number of operations on the tree. Example 2: Input: edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]], values = [20,10,9,7,4,3,5] Output: 40 Explanation: We can choose nodes 0, 2, 3, and 4. - The sum of values on the path from 0 to 4 is equal to 10. - The sum of values on the path from 0 to 3 is equal to 10. - The sum of values on the path from 0 to 5 is equal to 3. - The sum of values on the path from 0 to 6 is equal to 5. Therefore, the tree is healthy and the score is values[0] + values[2] + values[3] + values[4] = 40. It can be shown that 40 is the maximum score obtainable after any number of operations on the tree. Constraints: 2 <= n <= 2 * 104 edges.length == n - 1 edges[i].length == 2 0 <= ai, bi < n values.length == n 1 <= values[i] <= 109 The input is generated such that edges represents a valid tree.

Explanation

Here's the approach, complexity, and code for solving this problem:

• Dynamic Programming on Trees: We use a bottom-up dynamic programming approach, processing the tree from leaves to the root. The DP state at each node represents the maximum score achievable for the subtree rooted at that node.
• Two DP States: For each node, we maintain two DP states: one where the node's value is taken and one where it's not taken. We calculate these states by considering the corresponding states of the node's children and ensuring the path sums to leaf nodes are non-zero.

• Post-order Traversal: A post-order traversal is essential for calculating the DP states correctly since we must know the optimal scores for the subtrees rooted at the children before calculating the optimal score for the current node.

• Complexity: O(n) time and O(n) space, where n is the number of nodes.

Code

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

        dp = {}  # dp[node] = (score_take, score_not_take)

        def dfs(node, parent):
            dp[node] = (0, 0)  # Initialize both states to 0

            take = values[node]
            not_take = 0

            is_leaf = True
            for child in graph[node]:
                if child != parent:
                    is_leaf = False
                    dfs(child, node)

                    # Calculate 'take' state: We take current node's value
                    take += dp[child][1]  # Children must not take (to ensure path sum is values[node] > 0)

                    # Calculate 'not_take' state: We don't take current node's value
                    not_take += max(dp[child][0], dp[child][1]) # children can take or not take. we should add the maximum


            if is_leaf:  #leaf node
                dp[node] = (values[node], 0)
                return

            dp[node] = (take, not_take)
            return

        dfs(0, -1)
        return max(dp[0][0], dp[0][1])

    # Example Usage (from the prompt)
    edges1 = [[0,1],[0,2],[0,3],[2,4],[4,5]]
    values1 = [5,2,5,2,1,1]
    print(max_score_after_operations(edges1, values1)) # Output: 11

    edges2 = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]]
    values2 = [20,10,9,7,4,3,5]
    print(max_score_after_operations(edges2, values2)) # Output: 40


solve()