Problem Description

You have an undirected tree with n nodes labeled from 0 to n - 1, rooted at node 0. The tree structure is given by an array edges where edges[i] = [ai, bi] represents an edge between nodes ai and bi.

Each node i has a gate with an associated value in the amount array:

• If amount[i] is negative, it represents the cost to open the gate
• If amount[i] is positive or zero, it represents the reward for opening the gate

The game proceeds as follows:

Initial Setup:

• Alice starts at node 0 (the root)
• Bob starts at node bob (some other node in the tree)

Movement Rules:

• Every second, both Alice and Bob move to adjacent nodes simultaneously
• Alice moves toward a leaf node (a node with only one connection)
• Bob moves toward node 0 (the root)

Gate Opening Rules:

• When a player reaches a node, they interact with its gate
• If the gate hasn't been opened yet:
  • A single player pays the full cost or receives the full reward
  • If both players reach the node simultaneously, they split the cost/reward (each pays/receives amount[i] / 2)
• If the gate has already been opened, there's no cost or reward

Stopping Conditions:

• Alice stops when she reaches a leaf node
• Bob stops when he reaches node 0
• These events are independent (one can stop while the other continues)

Goal: Find the maximum net income Alice can achieve by choosing the optimal leaf node to travel to. Net income is the sum of all rewards minus all costs along her path.

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 and acyclic).

Is it a tree?

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

DFS

• Yes: We arrive at DFS as the recommended approach.

Conclusion: The flowchart suggests using DFS (Depth-First Search) for this tree problem.

This makes sense because:

1. We need to explore paths from the root to leaf nodes (Alice's movement)
2. We need to trace Bob's path from his starting node to the root
3. Tree traversal naturally fits the DFS pattern where we explore each branch fully before backtracking
4. The solution requires two DFS traversals:
   • First DFS: Track Bob's path and timing from node bob to node 0
   • Second DFS: Explore all possible paths Alice can take from node 0 to leaf nodes, calculating the maximum profit while considering Bob's timing at each node

Intuition

The key insight is that Bob's path is completely deterministic - he must travel from node bob to node 0 along the unique path that exists in a tree. This means we can precompute exactly when Bob will visit each node on his path.

Think of it this way: Alice has choices (she can pick which leaf to target), but Bob doesn't. Since Bob's movement is fixed, we can simulate his journey first and mark the time he reaches each node along his path.

Once we know Bob's timeline, Alice's problem becomes: "Given that certain nodes will be visited by Bob at specific times, what's the best leaf I can reach to maximize my profit?"

For each node Alice visits, three scenarios can occur:

1. Alice arrives before Bob - she gets the full amount[i]
2. Alice and Bob arrive simultaneously - they split amount[i] equally
3. Alice arrives after Bob - the gate is already open, so she gets nothing

This naturally leads to a two-phase approach:

Phase 1: Run DFS from Bob's starting position to find his path to node 0 and record the time he reaches each node on this path. Nodes not on Bob's path will never be visited by him, so we can mark them with an impossibly large time value.

Phase 2: Run DFS from Alice's starting position (node 0) to explore all possible paths to leaf nodes. At each node, compare Alice's arrival time with Bob's arrival time to determine how much profit/cost Alice incurs. Track the maximum profit among all leaf nodes Alice can reach.

The beauty of this approach is that it separates the deterministic part (Bob's movement) from the optimization part (Alice's choice), making the problem much more manageable.

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

Solution Approach

The solution implements the two-phase DFS approach we identified:

Phase 1: Track Bob's Path and Timing

The first DFS function dfs1(i, fa, t) traces Bob's path from his starting node to node 0:

• i is the current node
• fa is the parent node (to avoid revisiting)
• t is the current time

The function recursively explores neighbors until it finds node 0. When found, it returns True and backtracks, recording the time Bob reaches each node along his path in the ts array. The min operation ensures we keep the earliest arrival time if multiple paths exist (though in a tree, the path is unique).

Initially, we set ts[i] = n for all nodes (an impossibly large value), meaning Bob never visits them. After running dfs1, only nodes on Bob's path will have actual time values, with ts[bob] = 0 since Bob starts there.

Phase 2: Find Alice's Optimal Path

The second DFS function dfs2(i, fa, t, v) explores all possible paths Alice can take:

• i is the current node
• fa is the parent node
• t is Alice's current time
• v is Alice's accumulated profit/cost

At each node, we compare Alice's arrival time t with Bob's arrival time ts[i]:

• If t == ts[i]: They arrive simultaneously, so Alice gets amount[i] // 2
• If t < ts[i]: Alice arrives first, so she gets the full amount[i]
• If t > ts[i]: Bob already opened the gate, so Alice gets nothing (no change to v)

When Alice reaches a leaf node (identified by having only one neighbor, which is the parent), we update the global maximum ans with her accumulated value.

Data Structures Used:

• g: An adjacency list (defaultdict) to represent the tree structure
• ts: An array storing Bob's arrival time at each node
• ans: A global variable tracking the maximum profit Alice can achieve

The algorithm runs in O(n) time since each DFS visits every node once, making it efficient for tree traversal problems.

Example Walkthrough

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

Example Input:

• n = 4 nodes (labeled 0-3)
• edges = [[0,1], [1,2], [1,3]] (tree structure: 0 -- 1 -- 2, with 3 also connected to 1)
• amount = [5, -10, 8, 6] (gate values at each node)
• bob = 2 (Bob starts at node 2)

Tree Visualization:

    0 (5)
    |
    1 (-10)
   / \
  2   3
 (8) (6)

Phase 1: Track Bob's Path (from node 2 to node 0)

Bob must travel: 2 → 1 → 0

Running dfs1(2, -1, 0):

• Start at node 2 at time 0
• Move to node 1 at time 1
• Move to node 0 at time 2
• Backtrack and record times:
  • ts[2] = 0 (Bob starts here)
  • ts[1] = 1 (Bob reaches at second 1)
  • ts[0] = 2 (Bob reaches at second 2)
  • ts[3] = 4 (never visited, keeps default value)

Phase 2: Find Alice's Optimal Path (from node 0)

Alice has two possible leaf destinations: node 2 or node 3.

Path 1: Alice goes 0 → 1 → 2

• At node 0 (time 0): Bob arrives at time 2, so Alice arrives first → gets full amount[0] = 5
• At node 1 (time 1): Bob also arrives at time 1, simultaneous → gets amount[1]/2 = -10/2 = -5
• At node 2 (time 2): Bob was here at time 0, so gate already open → gets 0
• Total profit: 5 + (-5) + 0 = 0

Path 2: Alice goes 0 → 1 → 3

• At node 0 (time 0): Bob arrives at time 2, so Alice arrives first → gets full amount[0] = 5
• At node 1 (time 1): Bob also arrives at time 1, simultaneous → gets amount[1]/2 = -10/2 = -5
• At node 3 (time 2): Bob never visits (ts[3] = 4), so Alice arrives first → gets full amount[3] = 6
• Total profit: 5 + (-5) + 6 = 6

Result: Alice's maximum profit is 6 by choosing to go to leaf node 3.

This example demonstrates how:

1. Bob's path is fixed and deterministic (2→1→0)
2. We precompute when Bob reaches each node
3. Alice evaluates each possible leaf path, accounting for Bob's interference
4. The optimal choice maximizes Alice's net income

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm consists of two depth-first search (DFS) traversals:

• dfs1: Traverses from Bob's position to node 0, visiting each node at most once. This takes O(n) time.
• dfs2: Traverses from node 0 (Alice's starting position) to all leaf nodes, visiting each node exactly once. This also takes O(n) time.

Since both DFS operations run sequentially and each visits nodes at most once, the total time complexity is O(n) + O(n) = O(n).

Space Complexity: O(n)

The space complexity comes from:

• Graph adjacency list g: Stores all edges, using O(n) space (since it's a tree with n-1 edges, each edge is stored twice).
• Array ts: Stores timestamps for each node, requiring O(n) space.
• Recursive call stack: In the worst case (skewed tree), the recursion depth can be O(n).
• Other variables (ans, parameters) use O(1) space.

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

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

Common Pitfalls

1. Incorrectly Identifying Leaf Nodes

A critical pitfall is misidentifying leaf nodes in the tree. The current code checks if a node is a leaf using:

if len(graph[node]) == 1 and graph[node][0] == parent:

The Problem: This condition fails for node 0 (the root) when it has only one child. Since the root is called with parent = -1, the condition graph[node][0] == parent will be false even if node 0 has degree 1, causing it to not be recognized as a leaf.

Example Scenario:

• Tree with only 2 nodes: edges = [[0, 1]]
• Node 0 has degree 1 but won't be identified as a leaf
• Alice starts at node 0 (which is also a leaf in this case)

Solution:

def calculate_alice_max_profit(node: int, parent: int, time: int, current_profit: int) -> None:
    # Calculate profit at current node
    if time == bob_arrival_times[node]:
        current_profit += amount[node] // 2
    elif time < bob_arrival_times[node]:
        current_profit += amount[node]
  
    # Check if this is a leaf node (degree 1, but not the root with degree 1)
    is_leaf = len(graph[node]) == 1 and node != 0
    # Special case: root is also a leaf if it has degree 0 or 1
    if node == 0 and len(graph[node]) <= 1:
        is_leaf = True
  
    if is_leaf:
        nonlocal max_profit
        max_profit = max(max_profit, current_profit)
        return
  
    # Continue exploring children...

2. Integer Division for Negative Values

When Alice and Bob arrive simultaneously at a node with a negative amount (cost), using integer division // can produce unexpected results.

The Problem:

• Python's // operator rounds toward negative infinity
• For example: -5 // 2 = -3 (not -2.5 rounded to -2)
• This means Alice would pay MORE than half the cost

Example:

• amount[i] = -5
• Expected split: Each pays -2.5, rounded to -2
• Actual with //: Alice pays -3

Solution:

# Instead of:
current_profit += amount[node] // 2

# Use:
current_profit += amount[node] / 2  # Keep as float during calculation

# Or for integer arithmetic with correct rounding:
if amount[node] >= 0:
    current_profit += amount[node] // 2
else:
    # For negative values, use ceiling division to split costs fairly
    current_profit += -(-amount[node] // 2)  # This gives -2 for -5

3. Not Handling Bob's Path Correctly

The current implementation has a subtle bug in tracking Bob's arrival times.

The Problem: The code sets Bob's arrival time after finding his path:

find_bob_path_to_root(bob, -1, 0)
bob_arrival_times[bob] = 0  # Set after the DFS

But inside find_bob_path_to_root, when backtracking, it updates times for nodes on the path including Bob's starting position. This could lead to inconsistent timing.

Solution:

def find_bob_path_to_root(node: int, parent: int, time: int) -> bool:
    # Record Bob's time at this node first
    if node == bob:
        bob_arrival_times[node] = 0
  
    if node == 0:
        bob_arrival_times[node] = min(bob_arrival_times[node], time)
        return True
  
    for neighbor in graph[node]:
        if neighbor != parent and find_bob_path_to_root(neighbor, node, time + 1):
            # Only update if this isn't Bob's starting node
            if node != bob:
                bob_arrival_times[node] = min(bob_arrival_times[node], time)
            return True
  
    return False

# Then call without resetting bob's time afterward:
find_bob_path_to_root(bob, -1, 0)
# Remove: bob_arrival_times[bob] = 0

These pitfalls can cause incorrect results in edge cases, particularly with small trees, negative amounts, or when Bob's starting position has special properties.