Introduction

In this blog post, we will explore how to solve the LeetCode problem "2646" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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 and unrooted tree with n nodes indexed from 0 to n - 1. You are given the integer n and 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. Each node has an associated price. You are given an integer array price, where price[i] is the price of the ith node. The price sum of a given path is the sum of the prices of all nodes lying on that path. Additionally, you are given a 2D integer array trips, where trips[i] = [starti, endi] indicates that you start the ith trip from the node starti and travel to the node endi by any path you like. Before performing your first trip, you can choose some non-adjacent nodes and halve the prices. Return the minimum total price sum to perform all the given trips. Example 1: Input: n = 4, edges = [[0,1],[1,2],[1,3]], price = [2,2,10,6], trips = [[0,3],[2,1],[2,3]] Output: 23 Explanation: The diagram above denotes the tree after rooting it at node 2. The first part shows the initial tree and the second part shows the tree after choosing nodes 0, 2, and 3, and making their price half. For the 1st trip, we choose path [0,1,3]. The price sum of that path is 1 + 2 + 3 = 6. For the 2nd trip, we choose path [2,1]. The price sum of that path is 2 + 5 = 7. For the 3rd trip, we choose path [2,1,3]. The price sum of that path is 5 + 2 + 3 = 10. The total price sum of all trips is 6 + 7 + 10 = 23. It can be proven, that 23 is the minimum answer that we can achieve. Example 2: Input: n = 2, edges = [[0,1]], price = [2,2], trips = [[0,0]] Output: 1 Explanation: The diagram above denotes the tree after rooting it at node 0. The first part shows the initial tree and the second part shows the tree after choosing node 0, and making its price half. For the 1st trip, we choose path [0]. The price sum of that path is 1. The total price sum of all trips is 1. It can be proven, that 1 is the minimum answer that we can achieve. Constraints: 1 <= n <= 50 edges.length == n - 1 0 <= ai, bi <= n - 1 edges represents a valid tree. price.length == n price[i] is an even integer. 1 <= price[i] <= 1000 1 <= trips.length <= 100 0 <= starti, endi <= n - 1

Explanation

Here's the breakdown of the solution:

• Compute Trip Frequencies: Determine how many times each node is visited across all trips. This is crucial for deciding which nodes to halve the price of. We use Depth-First Search (DFS) to trace the paths for each trip.
• Dynamic Programming on Tree: Perform a dynamic programming traversal on the tree to calculate the minimum cost. For each node, consider two options: either halve the price of the node or don't. The decision depends on the frequency of visits from the first step and the minimum costs of the subtrees.

• Root the Tree and Optimize: The problem specifies an unrooted tree. Rooting the tree allows for easier DFS traversal. Arbitrarily, we can pick any node as the root (e.g., node 0).

• Runtime Complexity: O(n + m), where n is the number of nodes and m is the total length of all paths across all trips. In the worst case, each trip could visit every node, so m could be O(n * number of trips). Storage Complexity: O(n)

Code

    def minimum_price(n: int, edges: list[list[int]], price: list[int], trips: list[list[int]]) -> int:
    """
    Calculates the minimum total price sum to perform all given trips after halving prices of some non-adjacent nodes.

    Args:
        n: The number of nodes in the tree.
        edges: A list of edges representing the tree.
        price: A list of prices for each node.
        trips: A list of trips, where each trip is a pair of start and end nodes.

    Returns:
        The minimum total price sum.
    """

    graph = [[] for _ in range(n)]
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    counts = [0] * n

    def dfs(start, end, path, visited):
        if start == end:
            for node in path:
                counts[node] += 1
            return True

        visited[start] = True
        for neighbor in graph[start]:
            if not visited[neighbor]:
                if dfs(neighbor, end, path + [neighbor], visited):
                    return True
        return False

    for start, end in trips:
        visited = [False] * n
        dfs(start, end, [start], visited)

    dp = {}  # Memoization for DP

    def solve(node, parent):
        if (node, parent) in dp:
            return dp[(node, parent)]

        # Option 1: Don't halve the price of the current node
        cost1 = price[node] * counts[node]
        for neighbor in graph[node]:
            if neighbor != parent:
                cost1 += min(solve(neighbor, node))

        # Option 2: Halve the price of the current node
        cost2 = (price[node] // 2) * counts[node]
        for neighbor in graph[node]:
            if neighbor != parent:
                cost2 += solve(neighbor, node)[0]  # Must not halve the neighbors

        dp[(node, parent)] = (cost1, cost2)
        return cost1, cost2

    result = min(solve(0, -1))  # Start DFS from node 0 (arbitrary root)
    return result