LeetCode #2920 — HARD

Maximum Points After Collecting Coins From All Nodes

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 coins of its ancestors have been already collected.

Coins at node_i can be collected in one of the following ways:

Collect all the coins, but you will get coins[i] - k points. If coins[i] - k is negative then you will lose abs(coins[i] - k) points.
Collect all the coins, but you will get floor(coins[i] / 2) points. If this way is used, then for all the node_j present in the subtree of node_i, coins[j] will get reduced to floor(coins[j] / 2).

Return the maximum points you can get after collecting the coins from all the tree nodes.

Example 1:

Input: edges = [[0,1],[1,2],[2,3]], coins = [10,10,3,3], k = 5
Output: 11                        
Explanation: 
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
Explanation: 
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

Step 01

Brute Force Baseline

Problem summary: 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] = [ai, bi] indicates that there is an edge between nodes ai and bi 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 coins of its ancestors have been already collected. Coins at nodei can be collected in one of the following ways: Collect all the coins, but you will get coins[i] - k points. If coins[i] - k is negative then you will lose abs(coins[i] - k) points. Collect all the coins, but you will get floor(coins[i] / 2) points. If this way is used, then for all the nodej present in the subtree of nodei, coins[j] will get

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Bit Manipulation · Tree

Example 1

[[0,1],[1,2],[2,3]]
[10,10,3,3]
5

Example 2

[[0,1],[0,2]]
[8,4,4]
0

Step 02

Core Insight

What unlocks the optimal approach

Let <code>dp[x][t]</code> be the maximum points we can get from the subtree rooted at node <code>x</code> and the second operation has been used <code>t</code> times in its ancestors.
Note that the value of each <code>node <= 10<sup>4</sup></code>, so when <code>t >= 14</code> <code>dp[x][t]</code> is always <code>0</code>.
General equation will be: <code>dp[x][t] = max((coins[x] >> t) - k + sigma(dp[y][t]), (coins[x] >> (t + 1)) + sigma(dp[y][t + 1]))</code> where nodes denoted by <code>y</code> in the sigma, are the direct children of node <code>x</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2920: Maximum Points After Collecting Coins From All Nodes
class Solution {
    private int k;
    private int[] coins;
    private Integer[][] f;
    private List<Integer>[] g;

    public int maximumPoints(int[][] edges, int[] coins, int k) {
        this.k = k;
        this.coins = coins;
        int n = coins.length;
        f = new Integer[n][15];
        g = new List[n];
        Arrays.setAll(g, i -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        return dfs(0, -1, 0);
    }

    private int dfs(int i, int fa, int j) {
        if (f[i][j] != null) {
            return f[i][j];
        }
        int a = (coins[i] >> j) - k;
        int b = coins[i] >> (j + 1);
        for (int c : g[i]) {
            if (c != fa) {
                a += dfs(c, i, j);
                if (j < 14) {
                    b += dfs(c, i, j + 1);
                }
            }
        }
        return f[i][j] = Math.max(a, b);
    }
}

// Accepted solution for LeetCode #2920: Maximum Points After Collecting Coins From All Nodes
func maximumPoints(edges [][]int, coins []int, k int) int {
	n := len(coins)
	f := make([][]int, n)
	for i := range f {
		f[i] = make([]int, 15)
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	g := make([][]int, n)
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	var dfs func(int, int, int) int
	dfs = func(i, fa, j int) int {
		if f[i][j] != -1 {
			return f[i][j]
		}
		a := (coins[i] >> j) - k
		b := coins[i] >> (j + 1)
		for _, c := range g[i] {
			if c != fa {
				a += dfs(c, i, j)
				if j < 14 {
					b += dfs(c, i, j+1)
				}
			}
		}
		f[i][j] = max(a, b)
		return f[i][j]
	}
	return dfs(0, -1, 0)
}

# Accepted solution for LeetCode #2920: Maximum Points After Collecting Coins From All Nodes
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

// Accepted solution for LeetCode #2920: Maximum Points After Collecting Coins From All Nodes
/**
 * [2920] Maximum Points After Collecting Coins From All Nodes
 */
pub struct Solution {}

// submission codes start here
use std::cell::RefCell;

struct Search {
    memory: RefCell<Vec<Vec<i32>>>,
    children: Vec<Vec<usize>>,
    coins: Vec<i32>,
}

impl Search {
    fn new(children: Vec<Vec<usize>>, n: usize, coins: Vec<i32>) -> Self {
        Self {
            children,
            // 14是树可能的最大高度
            memory: RefCell::new(vec![vec![-1; 14]; n]),
            coins,
        }
    }

    fn search(&self, node: usize, parent: usize, f: usize, k: i32) -> i32 {
        if self.memory.borrow()[node][f] >= 0 {
            return self.memory.borrow()[node][f];
        }

        let mut result0 = (self.coins[node] >> f) - k;
        let mut result1 = self.coins[node] >> (f + 1);

        for &child in self.children[node].iter() {
            if child == parent {
                continue;
            }

            result0 += self.search(child, node, f, k);
            if f + 1 < 14 {
                result1 += self.search(child, node, f + 1, k);
            }
        }

        self.memory.borrow_mut()[node][f] = result0.max(result1);
        self.memory.borrow()[node][f]
    }
}

impl Solution {
    pub fn maximum_points(edges: Vec<Vec<i32>>, coins: Vec<i32>, k: i32) -> i32 {
        let n = coins.len();
        let mut children = vec![vec![]; n];

        for edge in edges {
            let (x, y) = (edge[0] as usize, edge[1] as usize);

            children[x].push(y);
            children[y].push(x);
        }

        let search = Search::new(children, n, coins);

        search.search(0, usize::MAX, 0, k)
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2920() {
        assert_eq!(
            11,
            Solution::maximum_points(
                vec![vec![0, 1], vec![1, 2], vec![2, 3]],
                vec![10, 10, 3, 3],
                5
            )
        );
    }
}

// Accepted solution for LeetCode #2920: Maximum Points After Collecting Coins From All Nodes
function maximumPoints(edges: number[][], coins: number[], k: number): number {
    const n = coins.length;
    const f: number[][] = Array.from({ length: n }, () => Array(15).fill(-1));
    const g: number[][] = Array.from({ length: n }, () => []);
    for (const [a, b] of edges) {
        g[a].push(b);
        g[b].push(a);
    }
    const dfs = (i: number, fa: number, j: number): number => {
        if (f[i][j] !== -1) {
            return f[i][j];
        }
        let a = (coins[i] >> j) - k;
        let b = coins[i] >> (j + 1);
        for (const c of g[i]) {
            if (c !== fa) {
                a += dfs(c, i, j);
                if (j < 14) {
                    b += dfs(c, i, j + 1);
                }
            }
        }
        return (f[i][j] = Math.max(a, b));
    };
    return dfs(0, -1, 0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log M)

Space

O(n × log M)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.