LeetCode #2973 — HARD

Find Number of Coins to Place in Tree Nodes

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

The Problem

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] = [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 integer array cost of length n, where cost[i] is the cost assigned to the i^th 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 <= a_i, b_i < n
cost.length == n
1 <= |cost[i]| <= 10⁴
The input is generated such that edges represents a valid tree.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

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

Pattern signal: Dynamic Programming · Tree

Example 1

[[0,1],[0,2],[0,3],[0,4],[0,5]]
[1,2,3,4,5,6]

Example 2

[[0,1],[0,2],[1,3],[1,4],[1,5],[2,6],[2,7],[2,8]]
[1,4,2,3,5,7,8,-4,2]

Example 3

[[0,1],[0,2]]
[1,2,-2]

Core Insight

What unlocks the optimal approach

Use DFS on the whole tree, for each subtree, save the largest three positive costs and the smallest three non-positive costs. This can be done by using two Heaps with the size of at most three.
You need to store at most six values at each subtree.
If there are more than three values in total, we can sort them. Let’s call the resultant array <code>A</code>, the maximum product of three is <code>max(A[0] * A[1] * A[n - 1], A[n - 1] * A[n - 2] * A[n - 3])</code>. Don’t forget to set the result to <code>0</code> if the value is negative.
If there are less than three values for a subtree, set its result to <code>1</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 #2973: Find Number of Coins to Place in Tree Nodes
class Solution {
    private int[] cost;
    private List<Integer>[] g;
    private long[] ans;

    public long[] placedCoins(int[][] edges, int[] cost) {
        int n = cost.length;
        this.cost = cost;
        ans = new long[n];
        g = new List[n];
        Arrays.fill(ans, 1);
        Arrays.setAll(g, i -> new ArrayList<>());
        for (int[] e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        dfs(0, -1);
        return ans;
    }

    private List<Integer> dfs(int a, int fa) {
        List<Integer> res = new ArrayList<>();
        res.add(cost[a]);
        for (int b : g[a]) {
            if (b != fa) {
                res.addAll(dfs(b, a));
            }
        }
        Collections.sort(res);
        int m = res.size();
        if (m >= 3) {
            long x = (long) res.get(m - 1) * res.get(m - 2) * res.get(m - 3);
            long y = (long) res.get(0) * res.get(1) * res.get(m - 1);
            ans[a] = Math.max(0, Math.max(x, y));
        }
        if (m >= 5) {
            res = List.of(res.get(0), res.get(1), res.get(m - 3), res.get(m - 2), res.get(m - 1));
        }
        return res;
    }
}

// Accepted solution for LeetCode #2973: Find Number of Coins to Place in Tree Nodes
func placedCoins(edges [][]int, cost []int) []int64 {
	n := len(cost)
	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)
	}
	ans := make([]int64, n)
	for i := range ans {
		ans[i] = int64(1)
	}
	var dfs func(a, fa int) []int
	dfs = func(a, fa int) []int {
		res := []int{cost[a]}
		for _, b := range g[a] {
			if b != fa {
				res = append(res, dfs(b, a)...)
			}
		}
		sort.Ints(res)
		m := len(res)
		if m >= 3 {
			x := res[m-1] * res[m-2] * res[m-3]
			y := res[0] * res[1] * res[m-1]
			ans[a] = max(0, int64(x), int64(y))
		}
		if m >= 5 {
			res = append(res[:2], res[m-3:]...)
		}
		return res
	}
	dfs(0, -1)
	return ans
}

# Accepted solution for LeetCode #2973: Find Number of Coins to Place in Tree Nodes
class Solution:
    def placedCoins(self, edges: List[List[int]], cost: List[int]) -> List[int]:
        def dfs(a: int, fa: int) -> List[int]:
            res = [cost[a]]
            for b in g[a]:
                if b != fa:
                    res.extend(dfs(b, a))
            res.sort()
            if len(res) >= 3:
                ans[a] = max(res[-3] * res[-2] * res[-1], res[0] * res[1] * res[-1], 0)
            if len(res) > 5:
                res = res[:2] + res[-3:]
            return res

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

// Accepted solution for LeetCode #2973: Find Number of Coins to Place in Tree Nodes
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2973: Find Number of Coins to Place in Tree Nodes
// class Solution {
//     private int[] cost;
//     private List<Integer>[] g;
//     private long[] ans;
// 
//     public long[] placedCoins(int[][] edges, int[] cost) {
//         int n = cost.length;
//         this.cost = cost;
//         ans = new long[n];
//         g = new List[n];
//         Arrays.fill(ans, 1);
//         Arrays.setAll(g, i -> new ArrayList<>());
//         for (int[] e : edges) {
//             int a = e[0], b = e[1];
//             g[a].add(b);
//             g[b].add(a);
//         }
//         dfs(0, -1);
//         return ans;
//     }
// 
//     private List<Integer> dfs(int a, int fa) {
//         List<Integer> res = new ArrayList<>();
//         res.add(cost[a]);
//         for (int b : g[a]) {
//             if (b != fa) {
//                 res.addAll(dfs(b, a));
//             }
//         }
//         Collections.sort(res);
//         int m = res.size();
//         if (m >= 3) {
//             long x = (long) res.get(m - 1) * res.get(m - 2) * res.get(m - 3);
//             long y = (long) res.get(0) * res.get(1) * res.get(m - 1);
//             ans[a] = Math.max(0, Math.max(x, y));
//         }
//         if (m >= 5) {
//             res = List.of(res.get(0), res.get(1), res.get(m - 3), res.get(m - 2), res.get(m - 1));
//         }
//         return res;
//     }
// }

// Accepted solution for LeetCode #2973: Find Number of Coins to Place in Tree Nodes
function placedCoins(edges: number[][], cost: number[]): number[] {
    const n = cost.length;
    const ans: number[] = Array(n).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 = (a: number, fa: number): number[] => {
        const res: number[] = [cost[a]];
        for (const b of g[a]) {
            if (b !== fa) {
                res.push(...dfs(b, a));
            }
        }
        res.sort((a, b) => a - b);
        const m = res.length;
        if (m >= 3) {
            const x = res[m - 1] * res[m - 2] * res[m - 3];
            const y = res[0] * res[1] * res[m - 1];
            ans[a] = Math.max(0, x, y);
        }
        if (m > 5) {
            res.splice(2, m - 5);
        }
        return res;
    };
    dfs(0, -1);
    return ans;
}

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 n)

Space

O(n)

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.

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.