There is 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 values of length n, where values[i] is the value associated with the ith node.
You start with a score of 0. In one operation, you can:
i.values[i] to your score.values[i] to 0.A tree is healthy if the sum of values on the path from the root to any leaf node is different than zero.
Return the maximum score you can obtain after performing these operations on the tree any number of times so that it remains healthy.
Example 1:
Input: edges = [[0,1],[0,2],[0,3],[2,4],[4,5]], values = [5,2,5,2,1,1] Output: 11 Explanation: We can choose nodes 1, 2, 3, 4, and 5. The value of the root is non-zero. Hence, the sum of values on the path from the root to any leaf is different than zero. Therefore, the tree is healthy and the score is values[1] + values[2] + values[3] + values[4] + values[5] = 11. It can be shown that 11 is the maximum score obtainable after any number of operations on the tree.
Example 2:
Input: edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]], values = [20,10,9,7,4,3,5] Output: 40 Explanation: We can choose nodes 0, 2, 3, and 4. - The sum of values on the path from 0 to 4 is equal to 10. - The sum of values on the path from 0 to 3 is equal to 10. - The sum of values on the path from 0 to 5 is equal to 3. - The sum of values on the path from 0 to 6 is equal to 5. Therefore, the tree is healthy and the score is values[0] + values[2] + values[3] + values[4] = 40. It can be shown that 40 is the maximum score obtainable after any number of operations on the tree.
Constraints:
2 <= n <= 2 * 104edges.length == n - 1edges[i].length == 20 <= ai, bi < nvalues.length == n1 <= values[i] <= 109edges represents a valid tree.When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
We need to find the best possible score by performing operations on a tree. The brute force approach tries every possible combination of operations on each node in the tree to see what score it would produce. Finally, it picks the combination that gives the highest score.
Here's how the algorithm would work step-by-step:
def maximum_score_after_applying_operations_on_a_tree_brute_force(node_values, edges):
number_of_nodes = len(node_values)
adjacency_list = [[] for _ in range(number_of_nodes)]
for edge_start, edge_end in edges:
adjacency_list[edge_start].append(edge_end)
adjacency_list[edge_end].append(edge_start)
maximum_score = 0
def calculate_score(operations):
current_score = 0
for node_index, operation_performed in enumerate(operations):
if operation_performed:
current_score += node_values[node_index]
return current_score
def explore_all_combinations(node_index, current_operations, visited):
nonlocal maximum_score
# Ensure we only process each node once
visited[node_index] = True
# Base case: If we've processed all nodes, calculate and update max score
if all(visited):
current_score = calculate_score(current_operations)
maximum_score = max(maximum_score, current_score)
return
# Recursive step: Explore performing operation or not
for next_node in range(number_of_nodes):
if not visited[next_node]:
# Explore WITH performing operation
current_operations[next_node] = True
explore_all_combinations(next_node, current_operations.copy(), visited.copy())
# Explore WITHOUT performing operation
current_operations[next_node] = False
explore_all_combinations(next_node, current_operations.copy(), visited.copy())
return
# Iterate to use each node as a starting point
for start_node in range(number_of_nodes):
# Represents if we perform the operation on the current node or not
initial_operations = [False] * number_of_nodes
# Keep track of which nodes have been visited
initial_visited = [False] * number_of_nodes
# Start the recursive exploration from each starting node
initial_operations[start_node] = True
explore_all_combinations(start_node, initial_operations.copy(), initial_visited.copy())
initial_operations[start_node] = False
explore_all_combinations(start_node, initial_operations.copy(), initial_visited.copy())
return maximum_scoreThe problem asks us to maximize a score on a tree by performing operations on its edges. Instead of brute-forcing, we can think of this as a dynamic programming problem where we work our way up from the leaves of the tree, making smart decisions at each step to guarantee the maximum possible score.
Here's how the algorithm would work step-by-step:
def maximum_score_after_operations(values, edges):
number_of_nodes = len(values)
adjacency_list = [[] for _ in range(number_of_nodes)]
for edge_start, edge_end in edges:
adjacency_list[edge_start].append(edge_end)
adjacency_list[edge_end].append(edge_start)
maximum_scores = {}
def calculate_max_score(node, parent):
if (node, parent) in maximum_scores:
return maximum_scores[(node, parent)]
# 'Activate' means including its value, skipping children.
activate_node = values[node]
for neighbor in adjacency_list[node]:
if neighbor != parent:
activate_node += calculate_max_score(neighbor, node)[1]
# 'Do not activate' means skipping its value, including children.
skip_node = 0
for neighbor in adjacency_list[node]:
if neighbor != parent:
# Choosing the maximum score between activation/skipping.
skip_node += max(calculate_max_score(neighbor, node))
maximum_scores[(node, parent)] = (activate_node, skip_node)
return maximum_scores[(node, parent)]
# Start from node 0 with no parent.
# Choose the larger of the activate and skip options.
result = max(calculate_max_score(0, -1))
return result| Case | How to Handle |
|---|---|
| Empty tree (no nodes) | Return 0 as the maximum score since there are no nodes to operate on. |
| Single node tree | Return the value of the single node, as it is the only possible score. |
| Tree with only two nodes | Compare the values of the two nodes and return the larger value or their sum depending on problem constraints. |
| Tree with all nodes having the same value | The algorithm should still function correctly, as the choice will be based on tree structure, not the diversity of values. |
| Tree with nodes having very large values (potential for integer overflow) | Use long long or appropriate large integer type to store intermediate sums/scores to prevent overflow. |
| Deeply skewed tree (linear chain) | This could lead to stack overflow if a recursive approach is used; use iterative DFS or BFS to avoid excessive recursion depth. |
| Tree with negative node values | The algorithm needs to consider negative values when making decisions about operations to maximize the overall score. |
| Tree structure prevents achieving optimal score based on individual node values | The algorithm should correctly handle cases where the optimal solution involves a complex combination of operations across the tree. |