2467. Most Profitable Path in a Tree - Detailed Explanation

Problem Statement

You are given a tree with n nodes (labeled from 0 to n-1) represented as an array of edges. The tree is rooted at node 0. Each node i has an associated integer value given by the array amount. In addition, there is an integer bob which indicates Bob’s starting node.

Gameplay:

• Alice starts at the root (node 0) and can choose any simple path from the root to a leaf.
• Bob starts at the given node bob and moves along the unique path from his starting node to the root.
• Both Alice and Bob move simultaneously at the same pace (one edge per time unit).

Profit Rules for Alice:

• For a node i:
  • If Alice reaches node i before Bob, she collects the full amount[i].
  • If Alice and Bob reach node i at the same time, she collects half of amount[i] (integer division).
  • If Bob reaches node i before Alice, she collects 0 from that node.

The objective is to choose a path for Alice from the root to a leaf such that the sum of the collected amounts is maximized.

Example 1

• Input:
  edges = [[0,1],[1,2],[1,3]]
bob = 3
amount = [-2, 4, 2, 6]

• Explanation:
  The tree structure is:

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

  Bob’s unique path from node 3 to 0 is: 3 → 1 → 0.

  • Bob’s times: node 3 at t=0, node 1 at t=1, node 0 at t=2.
    Consider the path 0 → 1 → 2 for Alice:
  • At node 0 (t=0): Bob arrives at t=2 → Alice collects -2.
  • At node 1 (t=1): Bob arrives at t=1 → same time → Alice collects 4/2 = 2.
  • At node 2 (t=2): Bob never visits → Alice collects 2.

  Total profit = -2 + 2 + 2 = 2.

• Output:
  2

Example 2

• Input:
  edges = [[0,1]]
bob = 1
amount = [5, 10]

• Explanation:
  The tree structure is:

      0 (5)
       |
      1 (10)
  

  Bob’s path: 1 → 0.

  • Bob’s times: node 1 at t=0, node 0 at t=1.
    For Alice’s only path 0 → 1:
  • At node 0 (t=0): Bob at t=1 → collect 5.
  • At node 1 (t=1): Bob at t=0 (Bob reached earlier) → collect 0.

  Total profit = 5 + 0 = 5.

• Output:
  5

Example 3

• Input:
  edges = [[0,1],[0,2],[1,3],[1,4]]
bob = 2
amount = [0, -5, -10, 20, 5]

• Explanation:
  The tree structure is:

          0 (0)
         /   \
      1 (-5)  2 (-10)
      /   \
   3 (20) 4 (5)
  

  Bob’s path: 2 → 0.

  • Bob’s times: node 2 at t=0, node 0 at t=1.

  Consider Alice’s path 0 → 1 → 3:

  • At node 0 (t=0): Bob at t=1 → collect 0.
  • At node 1 (t=1): Bob never visits → collect -5.
  • At node 3 (t=2): Bob never visits → collect 20.

  Total profit = 0 + (-5) + 20 = 15.

• Output:
  15

Constraints

• The number of nodes: 2 ≤ n ≤ 10⁵
• edges.length = n - 1
• 0 ≤ bob < n
• amount.length = n
• The tree is a connected acyclic graph.
• amount[i] can be negative, zero, or positive.

Hints

1. Trace Bob's Path:
   First, determine the unique path from Bob’s starting node to the root. Record the time at which Bob reaches each node along that path (starting at time 0 from Bob’s starting node).

2. Simulate Alice’s Journey:
   Perform a Depth-First Search (DFS) from the root. At each node, compare the time when Alice arrives with the time Bob reaches that node (if at all). Based on the comparison, decide how much profit Alice collects:

   • If Alice arrives earlier than Bob, take the full amount.
   • If they arrive at the same time, add half the amount.
   • If Alice arrives later, add nothing.

Approach 1: Brute Force DFS (Enumerating All Paths)

Idea:
A straightforward method is to enumerate every path from the root to each leaf using DFS. For each path, simulate the arrival times for Alice and compare with Bob’s precomputed arrival times on his path (if applicable). Then calculate the profit for that path and keep track of the maximum.

Limitations:

• Although DFS is used, the brute force nature is acceptable only because every node is visited once.
• However, if not optimized (for example, recomputing Bob’s time at each node), the solution can be inefficient.
• This approach works well for smaller trees but might be suboptimal for the maximum constraints if extra work is done at each node.

Approach 2: Optimal DFS with Precomputed Bob’s Times

Idea:

1. Precompute Bob’s Arrival Times:

   • Run a DFS (or BFS) from the root to record the parent of each node.
   • Use these parent pointers to reconstruct Bob’s unique path from his starting node back to the root.
   • For each node on Bob’s path, record the time (distance from Bob’s starting node). For nodes not on Bob’s path, consider Bob’s arrival time as infinity (i.e., Bob never reaches these nodes).

2. DFS for Alice’s Path:

   • Start DFS from the root (node 0) with an initial time of 0.
   • At each node during DFS, compare the current time (Alice’s arrival time) with Bob’s arrival time (from the precomputed values):
     • If Alice's time < Bob's time, add the full amount[node].
     • If Alice's time == Bob's time, add half of amount[node] (using integer division).
     • If Alice's time > Bob's time, add 0.
   • Continue the DFS to traverse all possible root-to-leaf paths, updating the running total profit and recording the maximum profit seen.

Benefits:

• This method avoids redundant computation by precomputing Bob’s times only once.
• The DFS ensures that every node is visited a single time, resulting in an efficient solution.

Complexity Analysis

• Time Complexity:

  • Building the graph: O(n)
  • DFS to compute parent pointers: O(n)
  • Reconstructing Bob’s path: O(h) where h is the height of the tree (worst-case O(n)).
  • DFS for Alice’s journey: O(n)
    Overall: O(n)

• Space Complexity:

  • Graph storage: O(n)
  • Parent and Bob’s time arrays: O(n)
  • DFS recursion stack: O(h) (worst-case O(n) for skewed trees)
    Overall: O(n)

Step-by-Step Walkthrough and Visual Example

1. Graph Construction:
   Convert the edge list into an adjacency list to represent the tree.

2. Parent Computation:
   Starting from node 0, perform DFS to record each node’s parent. This allows us to reconstruct Bob’s unique path back to the root.

3. Compute Bob’s Arrival Times:

   • Begin at Bob’s starting node.
   • Traverse up using the parent pointers.
   • For each node on Bob’s path, record the time (starting from 0 for Bob’s starting node, increasing by 1 as you move towards the root).

4. DFS for Alice’s Paths:

   • Start from node 0 with time = 0.
   • At each node, compare Alice’s current time with Bob’s arrival time.
   • Based on the comparison, add:
     • Full amount if Alice arrives first.
     • Half the amount if they arrive simultaneously.
     • Zero if Bob has already been there.
   • Continue until a leaf is reached, and update the maximum profit.

Visual Example:
Consider Example 1 again.

• Bob’s path: 3 (t=0) → 1 (t=1) → 0 (t=2)
• Alice’s DFS along path 0 → 1 → 2:
  • At node 0 (t=0): 0 < 2, add -2.
  • At node 1 (t=1): 1 == 1, add 4/2 = 2.
  • At node 2 (t=2): Bob never visits (t = ∞), add 2.
  • Total profit = -2 + 2 + 2 = 2.

Common Mistakes

• Not Precomputing Bob’s Path Correctly:
  Failing to correctly trace Bob’s unique path from his starting node to the root can lead to incorrect comparisons during DFS.

• Incorrectly Handling the Half Collection:
  When Alice and Bob arrive at the same time, ensure that the profit is exactly half (using integer division).

• Overlooking Negative Values:
  Some nodes might have negative amounts; always include these in your profit calculations as they affect the overall maximum.

Edge Cases

• Minimal Tree:
  A tree with only 2 nodes (the root and one child). The solution should correctly handle the simplest possible tree.

• Negative Profits on the Path:
  Ensure that the algorithm correctly handles cases where many nodes have negative values.

• Bob’s Starting Node is the Root:
  If Bob starts at node 0, then the profit at the root might be halved (if both arrive at the same time).

Alternative Variations

• Different Profit Rules:
  Instead of halving the amount when both arrive at the same time, one variation might subtract a fixed value. The DFS approach would need to adjust the profit calculation accordingly.

• Multiple Players:
  Introducing more players moving on different paths might require a multi-source BFS or simultaneous DFS.