Problem Description

You have an undirected tree with n nodes labeled from 0 to n - 1. Each node i has a value given by nums[i], where nums is a 0-indexed array of length n. The tree structure is defined by an array edges of length n - 1, where each edges[i] = [ai, bi] represents an edge between nodes ai and bi.

Your task is to delete some edges from the tree, which will split it into multiple connected components. The value of each component is defined as the sum of all nums[i] values for nodes i within that component.

The goal is to find the maximum number of edges you can delete such that all resulting connected components have the same value.

For example, if you delete k edges, you'll create k + 1 connected components. All these components must have equal sums of their node values.

The problem asks you to return the maximum number of edges that can be deleted while maintaining this equal-value constraint across all components.

Key points to understand:

• The tree initially has n nodes and n - 1 edges (as it's a tree)
• Deleting edges creates multiple subtrees (connected components)
• Each component's value is the sum of its nodes' values
• All components must have exactly the same total value
• You want to maximize the number of deleted edges (which maximizes the number of components)

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 involves an undirected tree with nodes and edges. A tree is a special type of graph (connected acyclic graph).

Is it a tree?

• Yes: The problem states we have an undirected tree with n nodes and n-1 edges, which is the definition of a tree structure.

DFS

• Yes: Since we're dealing with a tree structure, DFS (Depth-First Search) is the recommended approach.

Conclusion: The flowchart leads us directly to using DFS for this tree problem.

Why DFS is Appropriate

The DFS pattern fits perfectly for this problem because:

1. Tree Traversal: We need to traverse the tree to calculate subtree sums from bottom to top.

2. Subtree Value Calculation: DFS allows us to recursively compute the sum of values in each subtree. We can process children first, then aggregate their values at the parent node.

3. Component Detection: When we find a subtree whose sum equals our target value (s/k), we can "cut" it off by returning 0 to the parent, effectively simulating edge deletion.

4. Validation: DFS helps us validate whether a given partition scheme is feasible by checking if we can form components with equal values throughout the entire tree.

The solution uses DFS to explore different possibilities of dividing the tree into k components, where each component has value s/k. By traversing from leaves to root, we can determine which edges to cut to create the desired equal-valued components.

Intuition

Let's think about what happens when we delete edges. If we delete k-1 edges, we create exactly k connected components. For all components to have equal value, each component must have value total_sum / k.

The key insight is that total_sum / k must be an integer for a valid partition to exist. This immediately limits our choices for k - we can only consider values of k that evenly divide the total sum.

Now, how do we check if we can actually create k components with equal values? We need to verify that we can partition the tree into subtrees where each has exactly value target = total_sum / k.

Here's where DFS becomes crucial. Starting from any node (say node 0), we can traverse the tree and calculate subtree sums. When we encounter a subtree whose sum equals our target value, we know we can "cut" it off from its parent - this represents deleting an edge. We mark this by returning 0 to the parent, indicating this subtree has been separated.

But what if a subtree's sum exceeds the target? This means our current choice of k is invalid - we can't create equal-valued components with this target. We return -1 to signal failure.

To maximize the number of edges deleted (which means maximizing k), we try values of k from large to small. The first valid k we find gives us the maximum number of components possible, and thus the maximum number of edges we can delete is k-1.

There's also an optimization: the maximum possible k is bounded by min(n, total_sum / max_value), because each component must contain at least one node, and the smallest possible component value is at least the maximum node value in the tree.

The beauty of this approach is that DFS naturally handles the tree structure and allows us to make greedy decisions - whenever we find a valid subtree of the target size, we can immediately "cut" it, knowing this won't prevent us from finding other valid partitions in the remaining tree.

Learn more about Tree, Depth-First Search and Math patterns.

Solution Approach

Let's walk through the implementation step by step:

1. Graph Representation

First, we build an adjacency list representation of the tree using a defaultdict(list). For each edge [a, b], we add b to g[a] and a to g[b] since it's an undirected tree.

2. Enumeration of k (Number of Components)

We calculate:

• s = sum(nums): Total sum of all node values
• mx = max(nums): Maximum value among all nodes
• Starting k = min(n, s // mx): The maximum possible number of components

We iterate k from this maximum value down to 2 (since k=1 means no edges deleted). For each k, we check if s % k == 0. If not, we skip this k since we can't divide the sum evenly.

3. DFS Function for Validation

For each valid k, we set t = s // k as our target component value and use DFS to check if we can partition the tree:

def dfs(i, fa):
    x = nums[i]  # Start with current node's value
    for j in g[i]:
        if j != fa:  # Avoid going back to parent
            y = dfs(j, i)  # Get subtree sum from child
            if y == -1:
                return -1  # Propagate failure
            x += y  # Add child's contribution
    if x > t:
        return -1  # Subtree sum exceeds target - invalid
    return x if x < t else 0  # Return sum or 0 if we found a complete component

The DFS logic:

• Start with the current node's value
• Recursively calculate sums from all children (excluding parent to avoid cycles)
• If any child returns -1, propagate the failure upward
• Add valid child contributions to current subtree sum
• If subtree sum equals target t, return 0 (indicating this subtree can be cut off)
• If subtree sum is less than t, return the sum (this subtree needs to be combined with others)
• If subtree sum exceeds t, return -1 (invalid partition)

4. Finding the Answer

We start DFS from node 0 with parent -1. If dfs(0, -1) returns 0, it means we successfully partitioned the entire tree into k components of equal value. We return k - 1 as the number of edges deleted.

If no valid partition is found for any k, we return 0 (no edges can be deleted while maintaining equal component values).

Time Complexity: O(n * d) where d is the number of divisors of s. In the worst case, this could be O(n²).

Space Complexity: O(n) for the adjacency list and recursion stack.

Example Walkthrough

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

Example:

• nums = [6, 2, 2, 2, 6] (5 nodes with these values)
• edges = [[0,1], [1,2], [1,3], [3,4]]

The tree structure looks like:

    0(6)
     |
    1(2)
   /   \
  2(2)  3(2)
         |
        4(6)

Step 1: Calculate total sum and maximum value

• Total sum s = 6 + 2 + 2 + 2 + 6 = 18
• Maximum value mx = 6
• Maximum possible components k = min(5, 18//6) = min(5, 3) = 3

Step 2: Try k = 3 (attempting to create 3 components)

• Check if 18 % 3 == 0 ✓ (Yes, it divides evenly)
• Target value per component: t = 18/3 = 6

Step 3: Run DFS from node 0

Let's trace through the DFS:

1. Start at node 0 (value = 6)
   • Visit child 1
2. At node 1 (value = 2)
   • Visit child 2
     • Node 2 is a leaf with value 2
     • Returns 2 (less than target 6)
   • Visit child 3
     • At node 3 (value = 2)
       • Visit child 4
         • Node 4 is a leaf with value 6
         • Returns 0 (equals target, can be cut off!)
       • Node 3: x = 2 + 0 = 2
       • Returns 2 (less than target)
   • Node 1: x = 2 + 2 + 2 = 6
   • Returns 0 (equals target, can be cut off!)
3. Back at node 0: x = 6 + 0 = 6
   • Returns 0 (equals target)

Result: DFS returns 0, meaning we successfully created 3 components:

• Component 1: Node 0 alone (sum = 6)
• Component 2: Nodes 1, 2, 3 (sum = 2+2+2 = 6)
• Component 3: Node 4 alone (sum = 6)

We deleted 2 edges to create these 3 components.

Step 4: Check if we can do better with k = 2

• Target would be 18/2 = 9
• Running DFS would fail because no valid partition exists where each component sums to 9

Final Answer: Maximum edges we can delete = 2

The algorithm efficiently found that we can partition the tree into 3 equal-valued components by removing the edges between nodes 0-1 and nodes 3-4.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × √s)

The algorithm iterates through possible values of k from min(n, s // mx) down to 2. The key insight is that we're looking for divisors of s (the sum of all node values). The number of divisors of s is bounded by O(√s) in the average case, though it can be O(s^(1/3)) in the worst case for highly composite numbers. However, the practical bound used here is O(√s).

For each valid divisor (when s % k == 0), we perform a DFS traversal of the tree which takes O(n) time, where n is the number of nodes. The DFS visits each node exactly once and processes each edge twice (once from each direction).

Therefore, the total time complexity is O(number of divisors × DFS cost) = O(√s × n) = O(n × √s).

Space Complexity: O(n)

The space complexity consists of:

• The adjacency list g which stores all edges: O(n) space (since a tree with n nodes has n-1 edges)
• The recursive call stack for DFS which can go up to depth O(n) in the worst case (when the tree is a linear chain)
• Other variables like s, mx, t, k which use O(1) space

The dominant factor is the recursion stack and adjacency list, both of which are O(n), giving us a total space complexity of O(n).

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

Common Pitfalls

1. Incorrect Handling of Component Detection in DFS

A common mistake is misunderstanding when a subtree forms a complete component. Consider this incorrect implementation:

def dfs(node, parent):
    subtree_sum = nums[node]
    for neighbor in adjacency_list[node]:
        if neighbor != parent:
            child_sum = dfs(neighbor, node)
            if child_sum == -1:
                return -1
            subtree_sum += child_sum
  
    # WRONG: Only checking if sum equals target
    if subtree_sum == target_sum:
        return 0
    elif subtree_sum < target_sum:
        return subtree_sum
    else:
        return -1

Problem: This doesn't properly handle the case where we need to accumulate values from multiple subtrees before reaching the target sum.

Solution: The correct approach returns 0 when a complete component is found (allowing the parent to "cut" this subtree), returns the accumulated sum when it's less than target (to be combined with sibling subtrees), and returns -1 for invalid cases.

2. Off-by-One Error in Return Value

Many solutions incorrectly return num_components instead of num_components - 1:

# WRONG
if dfs(0, -1) == 0:
    return num_components  # This returns number of components, not edges deleted!

# CORRECT
if dfs(0, -1) == 0:
    return num_components - 1  # Number of edges deleted = components - 1

Problem: The question asks for the number of edges deleted, not the number of components created.

Solution: Remember that deleting k edges creates k + 1 components, so if we want num_components components, we need to delete num_components - 1 edges.

3. Inefficient Component Count Enumeration

Starting from k = 2 and going up is inefficient:

# INEFFICIENT
for num_components in range(2, num_nodes + 1):
    if total_sum % num_components == 0:
        target_sum = total_sum // num_components
        if dfs(0, -1) == 0:
            return num_components - 1

Problem: This approach finds the minimum number of edges to delete rather than the maximum. Also, it doesn't consider the constraint that each component must have at least one node.

Solution: Start from the maximum possible number of components and work downward. The maximum is bounded by min(n, total_sum // max_value) since each component needs at least the value of the largest node.

4. Missing Edge Case: Single Node Components

Not considering that the maximum component value constrains the minimum component sum:

# WRONG: Not considering node value constraints
max_components = num_nodes

# CORRECT: Each component must contain at least the max node value
max_components = min(num_nodes, total_sum // max(nums))

Problem: If we have a node with value 10 and total sum is 30, we can have at most 3 components (each with sum 10), not more.

Solution: Calculate the upper bound on components as min(n, total_sum // max(nums)) to ensure each component can accommodate at least the largest node value.

5. Graph Construction Error with Self-Loops

Be careful not to create duplicate edges or self-loops when building the adjacency list:

# POTENTIAL ISSUE if edges contain duplicates or self-loops
for node_a, node_b in edges:
    adjacency_list[node_a].append(node_b)
    adjacency_list[node_b].append(node_a)

Problem: While the problem guarantees a valid tree (no cycles, n-1 edges for n nodes), some implementations might not handle edge cases properly during testing.

Solution: The current implementation is correct for valid tree input. However, for robustness, you could add validation or use a set-based adjacency list if duplicate edges were possible.