# Solving Leetcode Interviews in Seconds with AI: Maximum Points After Collecting Coins From All Nodes


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2920" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> There exists an undirected tree rooted at node 0 with n nodes labeled from 0 to n - 1. 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 array coins of size n where coins[i] indicates the number of coins in the vertex i, and an integer k. Starting from the root, you have to collect all the coins such that the coins at a node can only be collected if the coins of its ancestors have been already collected. Coins at nodei can be collected in one of the following ways:  Collect all the coins, but you will get coins[i] - k points. If coins[i] - k is negative then you will lose abs(coins[i] - k) points. Collect all the coins, but you will get floor(coins[i] / 2) points. If this way is used, then for all the nodej present in the subtree of nodei, coins[j] will get reduced to floor(coins[j] / 2).  Return the maximum points you can get after collecting the coins from all the tree nodes.   Example 1:   Input: edges = [[0,1],[1,2],[2,3]], coins = [10,10,3,3], k = 5 Output: 11                         Explanation:  Collect all the coins from node 0 using the first way. Total points = 10 - 5 = 5. Collect all the coins from node 1 using the first way. Total points = 5 + (10 - 5) = 10. Collect all the coins from node 2 using the second way so coins left at node 3 will be floor(3 / 2) = 1. Total points = 10 + floor(3 / 2) = 11. Collect all the coins from node 3 using the second way. Total points = 11 + floor(1 / 2) = 11. It can be shown that the maximum points we can get after collecting coins from all the nodes is 11.   Example 2:    Input: edges = [[0,1],[0,2]], coins = [8,4,4], k = 0 Output: 16 Explanation:  Coins will be collected from all the nodes using the first way. Therefore, total points = (8 - 0) + (4 - 0) + (4 - 0) = 16.    Constraints:  n == coins.length 2 <= n <= 105 0 <= coins[i] <= 104 edges.length == n - 1 0 <= edges[i][0], edges[i][1] < n 0 <= k <= 104  

	# Explanation
	Here's a breakdown of the approach, followed by the Python code:

*   **Dynamic Programming on a Tree:** We use dynamic programming to explore both options for each node (collect with `coins[i] - k` or `floor(coins[i] / 2)`). The state is defined by the node index and the number of times the value has been divided by 2 due to applying the second option to an ancestor.
*   **Memoization:** The results of the subproblems are stored to avoid recomputation. This is crucial for efficiency.
*   **Depth-First Search (DFS):** The tree is traversed using DFS, allowing us to compute the optimal solution in a bottom-up manner.

*   **Runtime Complexity:** O(n * log(max(coins))), where n is the number of nodes, and log(max(coins)) is the maximum depth of the recursion which is limited by the number of times we can divide the coin values by 2 before they become 0.
*   **Storage Complexity:** O(n * log(max(coins))) due to the memoization table and recursion stack.

	
	# Code
	```python
	from functools import lru_cache

def maximum_points(edges, coins, k):
    n = len(coins)
    graph = [[] for _ in range(n)]
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    @lru_cache(None)
    def dfs(node, num_halvings, parent):
        """
        Calculates the maximum points obtainable from the subtree rooted at 'node',
        given that 'num_halvings' halvings have been applied by ancestors.
        """

        current_coin_value = coins[node] // (2 ** num_halvings)

        # Option 1: Collect using coins[i] - k
        option1 = current_coin_value - k
        for neighbor in graph[node]:
            if neighbor != parent:
                option1 += dfs(neighbor, num_halvings, node)

        # Option 2: Collect using floor(coins[i] / 2)
        option2 = current_coin_value // 2
        for neighbor in graph[node]:
            if neighbor != parent:
                option2 += dfs(neighbor, num_halvings + 1, node)

        return max(option1, option2)

    return dfs(0, 0, -1)
	```