What is the time complexity of Maximum Points After Collecting Coins From All Nodes?

The optimal solution has a time complexity of O(n * MAX_DIV).

What is the space complexity of Maximum Points After Collecting Coins From All Nodes?

The space complexity is O(n * MAX_DIV).

What data structures are used to solve Maximum Points After Collecting Coins From All Nodes?

This problem uses Array, Dynamic Programming, Bit Manipulation, Tree, Depth-First Search, Memoization. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Points After Collecting Coins From All Nodes?

Insight 1: Dynamic Programming (DP) is crucial because the decision at each node (collect all or half) impacts the points gained from the subtree. We need to explore all possible combinations of choices to maximize points. Insight 2: Tree traversal, specifically post-order traversal, is essential. We must process the children nodes before their parents, allowing the DP states to be correctly computed based on the optimal choices made in the subtrees.

Which companies ask LeetCode 2920. Maximum Points After Collecting Coins From All Nodes in interviews?

This problem is frequently asked at DE Shaw.

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Maximum Points After Collecting Coins From All Nodes - LeetCode 2920 Solution

Maximum Points After Collecting Coins From All Nodes - Complete Solution Guide

Maximum Points After Collecting Coins From All Nodes is LeetCode problem 2920, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There exists an undirected tree rooted at node 0 with n nodes labeled from 0 to n - 1 . You are given a 2D integer array edges of length n - 1 , where edges[i] = [a i , b i ] indicates that there is an edge between nodes a i and b i in the tree. You are also given a 0-indexed array coins of size n where coins[i] indicates the number of coins in the vertex i , and an integer k . Starting from the root, you have to collect all the coins such that the coins at a node can only be collected if the co

Detailed Explanation

The problem presents a tree structure where each node has a certain number of coins. Starting from the root, we need to collect coins from all nodes. Collecting coins from a node can be done in two ways: (1) collect all coins at the node and gain `coins[i] - k` points, or (2) collect coins and gain `floor(coins[i] / 2)` points. If the second way is used, then coins in the subtree of the current node are halved (floored). The goal is to find the maximum points that can be obtained by collecting coins from all the nodes in the tree. The coins at a node can only be collected if the coins of its ancestors have already been collected. The inputs are the edges of the tree, the number of coins at each node, and the constant 'k'. The output is the maximum points achievable.

Solution Approach

The solution utilizes dynamic programming with a post-order Depth-First Search (DFS) traversal of the tree. The `dp[u][d]` array stores the maximum points achievable from the subtree rooted at node `u` after 'd' divisions have been applied by the ancestors of `u`. The post-order traversal ensures that we compute the optimal points for the subtrees before computing the points for a node. For each node `u`, we consider two options: collecting coins with cost `k` or collecting coins by halving them. In the first option, the points are `coins[u] >> d - k` plus the maximum points from the children's subtrees without additional divisions. In the second option, the points are `coins[u] >> d / 2` plus the maximum points from the children's subtrees after applying an additional division (or reaching the maximum number of divisions). The algorithm returns `dp[0][0]`, which represents the maximum points achievable starting from the root node (node 0) with no initial divisions.

Step-by-Step Algorithm

Step 1: Build the Adjacency List: Create an adjacency list representation of the tree from the given `edges`.
Step 2: Construct Children List: Perform a Breadth-First Search (BFS) starting from the root (node 0) to determine the children of each node. This helps define the tree structure for the post-order traversal.
Step 3: Determine Post-Order Traversal: Generate a post-order traversal of the tree. This order is important because we must calculate the optimal score of each subtree *before* calculating the optimal score of its parent node.
Step 4: Initialize DP Table: Create a DP table `dp[n][MAX_DIV + 1]`, where `dp[u][d]` represents the maximum points achievable from the subtree rooted at node `u` with 'd' divisions already applied by its ancestors. Initialize all values to 0.
Step 5: Populate DP Table (Bottom-Up): Iterate through the nodes in the post-order traversal. For each node `u` and each division count `d` from 0 to `MAX_DIV`: calculate the points for both options: (1) collect with cost `k`, (2) collect with halving cost. The points gained from children are added in depending on the current choice at node `u`. Store the maximum of the two options in `dp[u][d]`.
Step 6: Return Result: Return `dp[0][0]`, which represents the maximum points achievable starting from the root with no initial divisions.

Key Insights

Insight 1: Dynamic Programming (DP) is crucial because the decision at each node (collect all or half) impacts the points gained from the subtree. We need to explore all possible combinations of choices to maximize points.
Insight 2: Tree traversal, specifically post-order traversal, is essential. We must process the children nodes before their parents, allowing the DP states to be correctly computed based on the optimal choices made in the subtrees.
Insight 3: Memoization is critical for efficiency. Since the same node and division count combinations will be encountered multiple times, storing computed results prevents redundant calculations and drastically improves performance.
Insight 4: The number of divisions is limited. Since coins[i] <= 10^4, any value will become 0 after approximately 14 divisions by 2. Therefore, we can limit the division states to a maximum of 14 (MAX_DIV) to keep complexity manageable.

Complexity Analysis

Time Complexity: O(n * MAX_DIV)

Space Complexity: O(n * MAX_DIV)

Topics

This problem involves: Array, Dynamic Programming, Bit Manipulation, Tree, Depth-First Search, Memoization.

Companies

Asked at: DE Shaw.

Maximum Points After Collecting Coins From All Nodes | LeetCode 2920

Hard

Array Dynamic Programming Bit Manipulation Tree Depth-First Search Memoization

Time: O(n * MAX_DIV)Space: O(n * MAX_DIV)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int maximumPoints(int[][] edges, int[] coins, int k) {
        int n = coins.length;
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.get(u).add(v);
            adj.get(v).add(u);
        }

        List<List<Integer>> children = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            children.add(new ArrayList<>());
        }
        Queue<Integer> q = new LinkedList<>();
        q.offer(0);
        Set<Integer> visited = new HashSet<>();
        visited.add(0);

        List<Integer> bfs_order = new ArrayList<>();
        int head = 0;
        while (head < q.size()) {
            int u = q.poll();
            bfs_order.add(u);
            for (int v : adj.get(u)) {
                if (!visited.contains(v)) {
                    visited.add(v);
                    children.get(u).add(v);
                    q.offer(v);
                }
            }
            
            q = new LinkedList<>(q); // Reinitialize q to avoid concurrent modification
        }

        Collections.reverse(bfs_order);
        List<Integer> post_order = bfs_order;

        // max(coins) is 10000. log2(10000) is approx 13.3.
        // After 14 divisions, any coin value becomes 0.
        int MAX_DIV = 14;

        // dp[u][d] = max points from subtree at u with d divisions from ancestors
        int[][] dp = new int[n][MAX_DIV + 1];

        for (int u : post_order) {
            for (int div = 0; div <= MAX_DIV; div++) {
                int current_coins = coins[u] >> div;

                // Option 1: Collect with cost k
                int option1_pts = current_coins - k;

                // Option 2: Collect with halving cost
                int option2_pts = current_coins / 2;

                for (int v : children.get(u)) {
                    option1_pts += dp[v][div];
                    option2_pts += dp[v][Math.min(div + 1, MAX_DIV)];
                }

                dp[u][div] = Math.max(option1_pts, option2_pts);
            }
        }

        return dp[0][0];
    }
}

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Problem Info

Description

Problem 2920: Maximum Points After Collecting Coins From All Nodes - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Bit Manipulation Tree Depth-First Search Memoization

Companies

DE Shaw

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2920

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Build the Adjacency List: Create an adjacency list representation of the tree from the given `edges`.
2
Step 2: Construct Children List: Perform a Breadth-First Search (BFS) starting from the root (node 0) to determine the children of each node. This helps define the tree structure for the post-order traversal.
3
Step 3: Determine Post-Order Traversal: Generate a post-order traversal of the tree. This order is important because we must calculate the optimal score of each subtree *before* calculating the optimal score of its parent node.
4
Step 4: Initialize DP Table: Create a DP table `dp[n][MAX_DIV + 1]`, where `dp[u][d]` represents the maximum points achievable from the subtree rooted at node `u` with 'd' divisions already applied by its ancestors. Initialize all values to 0.
5
Step 5: Populate DP Table (Bottom-Up): Iterate through the nodes in the post-order traversal. For each node `u` and each division count `d` from 0 to `MAX_DIV`: calculate the points for both options: (1) collect with cost `k`, (2) collect with halving cost. The points gained from children are added in depending on the current choice at node `u`. Store the maximum of the two options in `dp[u][d]`.
6
Step 6: Return Result: Return `dp[0][0]`, which represents the maximum points achievable starting from the root with no initial divisions.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n * MAX_DIV)

💾

Space Complexity

O(n * MAX_DIV)

📝

Why?

The time complexity is derived from the nested loops used to populate the DP table. The space complexity is due to the DP table and adjacency lists, which both scale linearly with the number of nodes.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

This approach uses a recursive function with memoization to explore all possible choices. The function takes the current node and the number of divisions as input and returns the maximum points achievable from the subtree rooted at that node. Memoization prevents redundant calculations.

Time: O(n * MAX_DIV)Space: O(n * MAX_DIV) + O(n) for recursion stack

Trade-offs: This approach is conceptually similar to the DP approach and has similar time and space complexities. The DP approach is usually slightly more efficient due to the iterative nature, avoiding function call overheads. The recursive memoization approach can be more intuitive to implement for some.

✨

Key Insights

1Insight 1: Dynamic Programming (DP) is crucial because the decision at each node (collect all or half) impacts the points gained from the subtree. We need to explore all possible combinations of choices to maximize points.
2Insight 2: Tree traversal, specifically post-order traversal, is essential. We must process the children nodes before their parents, allowing the DP states to be correctly computed based on the optimal choices made in the subtrees.
3Insight 3: Memoization is critical for efficiency. Since the same node and division count combinations will be encountered multiple times, storing computed results prevents redundant calculations and drastically improves performance.
4Insight 4: The number of divisions is limited. Since coins[i] <= 10^4, any value will become 0 after approximately 14 divisions by 2. Therefore, we can limit the division states to a maximum of 14 (MAX_DIV) to keep complexity manageable.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: k = 0. If k is 0, then always taking coins[i] - k is never a loss. The provided solution handles this correctly.
Edge case 2: coins[i] are small. If coins[i] are small, collecting `coins[i] / 2` might be beneficial even if `k` is small. The DP approach correctly handles this since it explores both options.
Edge case 3: Tree with only one node. The BFS will still create the 'children' data, however, it will be empty in this case. Thus, the DP will still operate correctly and return `coins[0] - k` since there are no children to consider.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not performing post-order traversal. If the tree traversal is not done in post-order, the DP states will not be correctly computed, leading to incorrect results.

✗Common mistake 2: Not limiting the number of divisions. If the number of divisions is not limited, the complexity can become much higher and may lead to Time Limit Exceeded errors. Since coins[i] becomes negligible after approx. 14 divisions, it is imperative to limit it to MAX_DIV

✗Common mistake 3: Not memoizing correctly. If memoization is not implemented, redundant calculations can lead to TLE.

✗Common mistake 4: Incorrectly calculating subtree points. Points accumulated must take into account *existing* number of divisions applied.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation in a hierarchical organization. Each node represents a department, and coins represents resources that need to be allocated. We can allocate all or half of the resources, affecting the downstream resources available. The DP helps optimize the allocation to maximize some overall company metric.

🏦Real-world scenario 2: Game strategy optimization in a game with a skill tree. The tree represents skills, the coins are experience points to invest, and k is the cost of unlocking each skill. A similar DP can be used to minimize XP to reach a certain endgame ability.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 11, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Points After Collecting Coins From All Nodes"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?