Maximum Points After Collecting Coins From All Nodes

Maximum Points After Collecting Coins From All Nodes

1. What is this problem about?

The Maximum Points After Collecting Coins From All Nodes interview question is an advanced tree-based optimization problem. You are given a tree with n nodes, where each node has a certain number of coins. You start at the root (node 0) and must collect coins from all nodes. For each node, you have two choices:

1. Collect coins normally: You get coins[i] - k points.
2. Collect coins with a reduction: You get floor(coins[i] / 2) points, but this also reduces the coins in all its descendant nodes by half.

The goal is to find the maximum total points you can collect from the entire tree. This problem introduces a global state change (the reduction) that affects subtrees, making it a complex recursive challenge.

2. Why is this asked in interviews?

This Maximum Points After Collecting Coins From All Nodes coding problem is often asked in high-level technical rounds at firms like DE Shaw. it tests your ability to handle recursive dependencies and state compression. The core challenge is realizing that since coins are halved, a node's coins can only be reduced a small number of times (about 14 times for typical coin limits) before they become zero. This observation is key to optimizing the solution and demonstrates your ability to identify constraints that simplify an otherwise exponential problem.

3. Algorithmic pattern used

The Array, Memoization, Depth-First Search, Bit Manipulation, Dynamic Programming, Tree interview pattern is essential here. You use DFS to traverse the tree and memoization to store results for a state defined by (current_node, number_of_reductions). Since the maximum value of a coin is usually around $10^4$ or $10^5$, after about 14 halvings, the value becomes 0. Therefore, the "number of reductions" state only needs to go up to 14. This transforms a potentially infinite state space into a manageable one.

4. Example explanation

Imagine a simple tree: Node 0 is the root, and it has one child, Node 1. k = 5. Node 0 has 20 coins, Node 1 has 20 coins.

Option A: No reduction at Node 0. Points at Node 0 = 20 - 5 = 15. Now consider Node 1 (no reduction inherited):

1. No reduction: 20 - 5 = 15. Total = 15 + 15 = 30.
2. Reduction: floor(20/2) = 10. Total = 15 + 10 = 25.

Option B: Reduction at Node 0. Points at Node 0 = floor(20/2) = 10. Now Node 1's coins are halved (effectively 10).

1. No reduction at Node 1: 10 - 5 = 5. Total = 10 + 5 = 15.
2. Reduction at Node 1: floor(10/2) = 5. Total = 10 + 5 = 15.

In this case, the maximum points are 30.

5. Common mistakes candidates make

A major mistake in the Maximum Points After Collecting Coins From All Nodes coding problem is not realizing the limited number of halvings. Without this insight, candidates might try to pass the full reduction factor or a huge state, leading to memory issues or TLE. Another error is not correctly propagating the reduction to all descendants. Candidates might only apply it to immediate children, which is incorrect. Finally, forgetting to handle the case where coins become zero after many halvings can lead to infinite recursion or incorrect base cases.

6. Interview preparation tip

When dealing with problems that involve repeated division or halving, always calculate the "logarithmic limit." Knowing how many times you can divide a number before it reaches a base case (like 0 or 1) is a common trick to reduce state space in Dynamic Programming. Practice tree DFS and ensure you are comfortable passing state down the recursion and returning aggregated values up.