Problem Description

You are given an undirected tree with n nodes labeled from 0 to n - 1, rooted at node 0. The tree structure is defined by a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates an edge between nodes ai and bi.

Each node has an associated cost given in a 0-indexed integer array cost of length n, where cost[i] represents the cost value of node i.

Your task is to determine how many coins to place at each node according to these rules:

1. If the subtree of node i contains fewer than 3 nodes: Place exactly 1 coin at node i.

2. If the subtree of node i contains 3 or more nodes:

   • Select 3 distinct nodes from the subtree of node i
   • Calculate the product of their cost values
   • Place coins equal to the maximum possible product from all combinations of 3 nodes
   • If this maximum product is negative, place 0 coins instead

The subtree of a node includes the node itself and all its descendants.

Return an array coin of size n where coin[i] represents the number of coins to be placed at node i.

Example considerations:

• When calculating the maximum product of 3 nodes, you need to consider that negative costs can exist
• The maximum product might come from 3 positive values (largest three) or from 2 negative values and 1 positive value (two most negative × one most positive)
• The tree is undirected but rooted at node 0, meaning we traverse it as a rooted tree structure

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem explicitly states we have an undirected tree with nodes and edges. A tree is a special type of graph (connected and acyclic).

Is it a tree?

• Yes: The problem clearly states we're working with an undirected tree with n nodes and n-1 edges, rooted at node 0.

DFS

• Yes: Since we identified this as a tree problem, the flowchart directs us to use DFS (Depth-First Search).

Why DFS is appropriate for this problem:

• We need to process each node's subtree to calculate the number of coins to place
• For each node, we need information about all nodes in its subtree (their cost values)
• DFS naturally processes subtrees bottom-up, allowing us to collect information from children before processing the parent
• The recursive nature of DFS perfectly matches the recursive structure of calculating subtree properties

Conclusion: The flowchart correctly identifies this as a tree problem and suggests using DFS. This aligns perfectly with the solution approach, where we use DFS to traverse the tree, collect cost values from each subtree, and calculate the maximum product of 3 nodes' costs for each subtree to determine the number of coins to place at each node.

Intuition

When we need to calculate something for each node based on its subtree, DFS is the natural choice because it processes children before parents, allowing us to aggregate information bottom-up.

The key insight is recognizing what information we actually need from each subtree. For any node, we need to find the maximum product of 3 distinct cost values from its subtree.

At first, you might think we need to keep track of all cost values in each subtree, but that would be inefficient. Instead, let's think about which values could possibly contribute to the maximum product:

1. If all costs are positive: The maximum product comes from the 3 largest values.
2. If we have negative costs: The maximum could come from either:
   • The 3 largest positive values (if they exist)
   • 2 smallest (most negative) values × 1 largest positive value (negative × negative = positive)

This means we don't need all values - we only need to track the 2 smallest and 3 largest values from each subtree. This is a crucial optimization.

The DFS approach naturally builds up this information:

1. Start at a node and recursively visit all children
2. Collect the important cost values (smallest 2 and largest 3) from each child's subtree
3. Add the current node's cost to this collection
4. Sort to identify the smallest and largest values
5. Calculate the maximum product for this node using the formula: max(0, largest3 × largest2 × largest1, smallest1 × smallest2 × largest1)
6. Pass up only the 2 smallest and 3 largest values to the parent (at most 5 values total)

This pruning strategy (keeping only 5 values) ensures that even as we traverse up the tree, we maintain constant space per recursive call while still having all the information needed to calculate the maximum product at any ancestor node.

Learn more about Tree, Depth-First Search, Dynamic Programming, Sorting and Heap (Priority Queue) patterns.

Solution Approach

The implementation follows a DFS pattern with optimization for maintaining only the necessary cost values from each subtree.

Data Structure Setup:

• Build an adjacency list g where g[a] contains all neighbor nodes of node a
• Initialize answer array ans with all values set to 1 (default for subtrees with less than 3 nodes)

Core DFS Function: The dfs(a, fa) function traverses the tree and returns an array containing the most relevant cost values from the subtree rooted at node a:

1. Initialize result array: Start with res = [cost[a]] containing the current node's cost

2. Traverse children: For each neighbor b of node a:

   • Skip if b is the parent node fa (to avoid going back up the tree)
   • Recursively call dfs(b, a) and extend res with the returned values

3. Sort the collected costs: After collecting all costs from the subtree, sort res to easily identify smallest and largest values

4. Calculate coins for current node:

   • If len(res) >= 3: Calculate the maximum product using two possible combinations:
     • res[-3] × res[-2] × res[-1]: Product of 3 largest values
     • res[0] × res[1] × res[-1]: Product of 2 smallest and 1 largest value
     • Take max(0, combination1, combination2) to handle negative products
   • Otherwise: Keep the default value of 1 coin

5. Optimize the return values: If len(res) > 5, prune the array to keep only:

   • res[:2]: The 2 smallest values
   • res[-3:]: The 3 largest values

   This ensures we pass at most 5 values up the tree while maintaining all necessary information for parent calculations

Algorithm Execution:

• Start DFS from root node 0 with parent -1: dfs(0, -1)
• The function populates the ans array during traversal
• Return the completed ans array

Time Complexity: O(n × log n) where n is the number of nodes, due to sorting at each node (but each value participates in at most O(log n) sorts as it moves up the tree)

Space Complexity: O(n) for the adjacency list and recursion stack, with each recursive call maintaining at most 5 cost values

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example Tree:

edges = [[0,1], [0,2], [2,3], [2,4]]
cost = [2, -1, 3, 5, -2]

Tree structure:
      0(2)
     / \
   1(-1) 2(3)
        / \
      3(5) 4(-2)

Step-by-step DFS traversal:

1. Start at node 0, visit child 1:

   • Node 1 is a leaf, subtree has only 1 node
   • res = [-1], length < 3, so ans[1] = 1
   • Return [-1] to parent

2. From node 0, visit child 2:

   2a. From node 2, visit child 3:

   • Node 3 is a leaf
   • res = [5], length < 3, so ans[3] = 1
   • Return [5] to parent

   2b. From node 2, visit child 4:

   • Node 4 is a leaf
   • res = [-2], length < 3, so ans[4] = 1
   • Return [-2] to parent

   2c. Back at node 2:

   • Collect costs from subtree: res = [3] (node 2's cost)
   • Add child 3's values: res = [3, 5]
   • Add child 4's values: res = [3, 5, -2]
   • Sort: res = [-2, 3, 5]
   • Length = 3, calculate products:
     • Three largest: 3 × 5 × (-2) = -30
     • Two smallest × one largest: (-2) × 3 × 5 = -30
     • ans[2] = max(0, -30, -30) = 0
   • Return [-2, 3, 5] to parent

3. Back at node 0:

   • Collect costs from subtree: res = [2] (node 0's cost)
   • Add child 1's values: res = [2, -1]
   • Add child 2's values: res = [2, -1, -2, 3, 5]
   • Sort: res = [-2, -1, 2, 3, 5]
   • Length = 5, calculate products:
     • Three largest: 2 × 3 × 5 = 30
     • Two smallest × one largest: (-2) × (-1) × 5 = 10
     • ans[0] = max(0, 30, 10) = 30
   • Since this is the root, we don't need to return values

Final Result: ans = [30, 1, 0, 1, 1]

Key Observations:

• Leaf nodes (1, 3, 4) have subtrees with < 3 nodes, so they get 1 coin
• Node 2's subtree has exactly 3 nodes, but the maximum product is negative, so it gets 0 coins
• Node 0's subtree has 5 nodes, and the maximum product (2 × 3 × 5 = 30) comes from the three largest positive values
• The optimization of keeping only 5 values (2 smallest + 3 largest) when returning from node 2 would still preserve all necessary information for calculating node 0's result

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n)

The algorithm performs a depth-first search (DFS) traversal of the tree starting from node 0. During the traversal:

• Each node is visited exactly once, contributing O(n) to the traversal itself
• At each node, we collect costs from all descendants in its subtree and sort them
• The sorting operation at each node takes O(k × log k) where k is the size of the subtree rooted at that node
• In the worst case (a linear tree), the root node would need to sort all n elements, giving O(n × log n)
• While multiple sorts occur across different nodes, the amortized analysis shows that each element participates in sorting operations along its path to the root
• The dominant operation is the sorting at nodes with large subtrees, particularly near the root

Therefore, the overall time complexity is O(n × log n).

Space Complexity: O(n)

The space usage consists of:

• The adjacency list g which stores all edges: O(n) (since it's a tree with n-1 edges)
• The answer array ans: O(n)
• The recursion call stack depth: O(n) in the worst case (linear tree)
• The res list at each recursive call: Although we collect costs from subtrees, we trim the list to at most 5 elements (res = res[:2] + res[-3:]), so each call maintains at most O(1) additional space for the trimmed result
• The total space for all res lists across the recursion is bounded by O(n)

Therefore, the overall space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Forgetting to Handle Negative Products

Issue: A common mistake is only considering the product of the three largest values when calculating the maximum product. This fails when negative values are present in the cost array.

Why it happens: With negative costs, the maximum product might come from multiplying two negative values (which gives a positive) with one positive value. For example, if costs are [-10, -5, 1, 2, 3], the maximum product is (-10) × (-5) × 3 = 150, not 1 × 2 × 3 = 6.

Solution: Always check both combinations:

# Incorrect approach
max_product = subtree_costs[-3] * subtree_costs[-2] * subtree_costs[-1]

# Correct approach
max_product = max(
    subtree_costs[-3] * subtree_costs[-2] * subtree_costs[-1],  # 3 largest
    subtree_costs[0] * subtree_costs[1] * subtree_costs[-1],    # 2 smallest, 1 largest
    0  # Handle negative products
)

Pitfall 2: Not Optimizing the Returned Cost Array

Issue: Returning all collected costs from each subtree without pruning leads to exponential growth in the size of arrays being passed up the tree, causing time and memory limit exceeded errors.

Why it happens: In a deep tree, without pruning, each node would accumulate all costs from its entire subtree. A parent node would then need to sort and process increasingly large arrays.

Solution: After sorting, keep only the values that matter for future calculations:

# Inefficient - returns entire subtree
return subtree_costs

# Efficient - returns only necessary values
if len(subtree_costs) > 5:
    subtree_costs = subtree_costs[:2] + subtree_costs[-3:]
return subtree_costs

Pitfall 3: Incorrect Tree Traversal

Issue: Treating the undirected edges as a directed tree without properly tracking the parent node, leading to infinite recursion or incorrect subtree calculations.

Why it happens: The edges represent an undirected graph, but we need to traverse it as a rooted tree. Without tracking the parent, DFS might revisit nodes.

Solution: Pass the parent node in DFS and skip it during traversal:

def dfs(node, parent):
    for child in adjacency_list[node]:
        if child != parent:  # Critical check to avoid going back to parent
            child_costs = dfs(child, node)

Pitfall 4: Off-by-One Errors in Array Indexing

Issue: Using incorrect indices when accessing the smallest or largest values after sorting, especially when the array has exactly 3 elements.

Why it happens: Confusion about zero-based indexing and negative indexing in Python.

Solution: Remember that for a sorted array:

• arr[0] and arr[1] are the two smallest values
• arr[-1], arr[-2], and arr[-3] are the three largest values (in ascending order)

# For array [1, 2, 3, 4, 5]:
# arr[0] = 1 (smallest)
# arr[1] = 2 (second smallest)
# arr[-3] = 3 (third largest)
# arr[-2] = 4 (second largest)
# arr[-1] = 5 (largest)