#2920

Hard

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Maximum Points After Collecting Coins From All Nodes

Find the maximum points after collecting coins from all nodes of a tree using binary-tree traversal and state tracking.

array dynamic programming bit manipulation tree depth first search

Problem Statement

You are given an undirected tree with n nodes labeled from 0 to n-1. The tree is rooted at node 0, and there are n-1 edges defined in the 2D array edges. Each node contains a certain number of coins, given by the array coins. Starting from the root, you must collect all coins while ensuring coins at a node can only be collected if the coins of its ancestors are already collected. You are also given an integer k which affects the coin collection method.

Coins can be collected in one of two ways. In the first way, all coins are collected from a node, reducing the total points by k for each coin. In the second way, only half of the coins are collected, and the remaining coins at a node are reduced by half. The goal is to maximize the total points after collecting coins from all nodes in the tree.

Examples

Example 1

Input: edges = [[0,1],[1,2],[2,3]], coins = [10,10,3,3], k = 5

Output: 11

Collect all the coins from node 0 using the first way. Total points = 10 - 5 = 5. Collect all the coins from node 1 using the first way. Total points = 5 + (10 - 5) = 10. Collect all the coins from node 2 using the second way so coins left at node 3 will be floor(3 / 2) = 1. Total points = 10 + floor(3 / 2) = 11. Collect all the coins from node 3 using the second way. Total points = 11 + floor(1 / 2) = 11. It can be shown that the maximum points we can get after collecting coins from all the nodes is 11.

Example 2

Input: edges = [[0,1],[0,2]], coins = [8,4,4], k = 0

Output: 16

Coins will be collected from all the nodes using the first way. Therefore, total points = (8 - 0) + (4 - 0) + (4 - 0) = 16.

Constraints

n == coins.length
2 <= n <= 105
0 <= coins[i] <= 104
edges.length == n - 1
0 <= edges[i][0], edges[i][1] < n
0 <= k <= 104

Solution Approach

Binary-Tree Traversal with Dynamic Programming

Start by performing a depth-first search (DFS) traversal from the root node. At each node, calculate the maximum points based on the state of the subtree rooted at that node. Use dynamic programming to store intermediate results for optimal performance. For each node, track the points based on both coin collection methods.

Memoization for Efficiency

To avoid redundant calculations, use memoization to store results of subproblems. This helps in reusing the already computed results for each node's subtree, significantly reducing time complexity. Memoization allows for efficient recursive traversal while considering the state changes due to different coin collection methods.

Tracking State with dp

Define a state dp[x][t] representing the maximum points for the subtree rooted at node x where t indicates how many times the second collection method has been used. Track both collection methods at each node and calculate the maximum points at each step based on the state of the node and its ancestors.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the traversal and the number of subproblems to be solved, typically O(n). Space complexity depends on the memoization storage, also O(n). The complexity can vary depending on the chosen approach for dynamic programming and state tracking.

What Interviewers Usually Probe

Does the candidate efficiently use dynamic programming and memoization to avoid redundant calculations?
Can the candidate demonstrate understanding of how state tracking helps maximize the points for this problem?
Does the candidate effectively use depth-first search for tree traversal while considering the two coin collection methods?

Common Pitfalls or Variants

Common pitfalls

Failing to implement the second coin collection method properly and not adjusting the remaining coins correctly.
Not using memoization or dynamic programming to store intermediate results, leading to inefficient solutions.
Incorrectly handling the state transitions, especially when switching between the two coin collection methods.

Follow-up variants

Variation 1: What if coins are collected in a specific order or with additional constraints?
Variation 2: How does the problem change if the tree is not binary, or if nodes can have more than two children?
Variation 3: What if you had to collect coins starting from multiple nodes simultaneously?

FAQ

What is the core pattern in the Maximum Points After Collecting Coins From All Nodes problem?

The core pattern involves binary-tree traversal combined with dynamic programming and state tracking to maximize the total points from coin collection.

How can dynamic programming be applied to this problem?

Dynamic programming is used to store intermediate results for subproblems, allowing the efficient calculation of maximum points from each subtree while tracking the state transitions.

What are the key challenges in solving Maximum Points After Collecting Coins From All Nodes?

Key challenges include handling both coin collection methods efficiently and using memoization to avoid redundant calculations.

What does the second coin collection method do in this problem?

The second method allows for half of the coins to be collected, reducing the total but leaving coins in the node that are halved for further collection.

How does GhostInterview assist in solving this problem?

GhostInterview breaks down the problem into digestible steps, providing insights on dynamic programming, state tracking, and efficient tree traversal.

terminal

Solution

Solution 1: Memoization Search

First, we construct a graph $g$ based on the edges given in the problem, where $g[i]$ represents all adjacent nodes of node $i$. Then we can use the method of memoization search to solve this problem.

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class Solution:
    def maximumPoints(self, edges: List[List[int]], coins: List[int], k: int) -> int:
        @cache
        def dfs(i: int, fa: int, j: int) -> int:
            a = (coins[i] >> j) - k
            b = coins[i] >> (j + 1)
            for c in g[i]:
                if c != fa:
                    a += dfs(c, i, j)
                    if j < 14:
                        b += dfs(c, i, j + 1)
            return max(a, b)

        n = len(coins)
        g = [[] for _ in range(n)]
        for a, b in edges:
            g[a].append(b)
            g[b].append(a)
        ans = dfs(0, -1, 0)
        dfs.cache_clear()
        return ans

Continue Practicing

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