How do you solve Find Number of Coins to Place in Tree Nodes on LeetCode?

Find Number of Coins to Place in Tree Nodes (LeetCode #2973) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using heaps allows efficient tracking of top costs.

What is the brute force solution for Find Number of Coins to Place in Tree Nodes?

The brute force approach for Find Number of Coins to Place in Tree Nodes is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will calculate the maximum product of costs for each subtree by traversing the entire tree for each node. This means for each node, we will gather all costs in its subtree and compute the maximum product of any three distinct costs. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Find Number of Coins to Place in Tree Nodes?

Explain your thought process clearly. Discuss edge cases like negative costs or small trees. Practice DFS and tree traversal problems.

What are common mistakes when solving Find Number of Coins to Place in Tree Nodes?

Not handling negative costs correctly. Forgetting to check subtree sizes before calculating products.

#2973

Find Number of Coins to Place in Tree Nodes

Hard

Dynamic ProgrammingTreeDepth-First SearchSortingHeap (Priority Queue)Depth-First SearchHeapsDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

In the optimal approach, we will use Depth-First Search (DFS) while maintaining two heaps to efficiently track the top three positive costs and the bottom three non-positive costs in each subtree. This allows us to compute the maximum product in linear time.

⚙️

Algorithm

3 steps

1Step 1: Create a graph representation of the tree using adjacency lists.
2Step 2: Perform a DFS traversal starting from the root node, maintaining two heaps for costs.
3Step 3: For each node, if its subtree size is less than 3, place 1 coin. Otherwise, calculate the maximum product using the top three positive costs from the heaps.

solution.py38 lines

1import heapq
2from collections import defaultdict
3
4def findCoins(edges, cost):
5    graph = defaultdict(list)
6    for a, b in edges:
7        graph[a].append(b)
8        graph[b].append(a)
9
10    def dfs(node, parent):
11        pos_heap = []
12        neg_heap = []
13        for neighbor in graph[node]:
14            if neighbor != parent:
15                p, n = dfs(neighbor, node)
16                pos_heap += p
17                neg_heap += n
18        pos_heap.append(cost[node])
19        pos_heap.sort(reverse=True)
20        neg_heap.sort()
21
22        if len(pos_heap) > 3:
23            pos_heap = pos_heap[:3]
24        if len(neg_heap) > 3:
25            neg_heap = neg_heap[:3]
26
27        max_product = 0
28        if len(pos_heap) == 3:
29            max_product = pos_heap[0] * pos_heap[1] * pos_heap[2]
30        result[node] = max(max_product, 0)
31        return pos_heap, neg_heap
32
33    result = [0] * len(cost)
34    dfs(0, -1)
35    for i in range(len(result)):
36        if result[i] == 0:
37            result[i] = 1
38    return result

ℹ

Complexity note: This complexity is linear because we traverse each node once and maintain a limited number of costs in heaps, leading to efficient calculations.

1Using heaps allows efficient tracking of top costs.
2DFS is a natural fit for tree traversal.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.