Leetcode #2973: Find Number of Coins to Place in Tree Nodes

In this guide, we solve Leetcode #2973 Find Number of Coins to Place in Tree Nodes in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.

Quick Facts

Difficulty: Hard
Premium: No
Tags: Tree, Depth-First Search, Dynamic Programming, Sorting, Heap (Priority Queue)

Intuition

The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.

A carefully chosen DP state captures exactly what we need to build the final answer.

Approach

Define the DP state and recurrence, then compute states in the correct order.

Optionally compress space once the recurrence is clear.

Steps:

Choose a DP state definition.
Write the recurrence and base cases.
Compute states in the correct order.

Example

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.

Python Solution

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

Complexity

The time complexity is $O(n \times \log n)$ , and the space complexity is $O(n)$ . The space complexity is $O(n)$ .

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Python Solution

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

Leetcode #2973: Find Number of Coins to Place in Tree Nodes

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #2973: Find Number of Coins to Place in Tree Nodes

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview