Problem Description

In this problem, we are tasked with finding the minimum total price sum of trips across an undirected tree. The tree consists of n nodes, each with an associated price. Connections between the nodes are described by the edges list, where each edge is a connection between two nodes.

For the trips, we have a list of pairs indicating starting and ending nodes for each trip. A unique aspect of this problem is the ability to halve the prices of some non-adjacent nodes before starting any trips. This presents an optimization problem where we need to decide which node prices to reduce in order to minimize the total cost of all trips.

The goal is to determine the minimum possible sum of prices for all trips.

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 scenario involves various flights connecting different cities, which can be represented as a graph where each city is a node and each flight is a directed edge with possibly varying costs.

Is it a tree?

• No: As the graph includes flights between many cities and it could include cycles (a city can have multiple incoming and outgoing flights), it's not a tree structure.

Is the problem related to directed acyclic graphs (DAGs)?

• No: While directed, the graph likely contains cycles due to the nature of flight routes between cities, ruling out DAG.

Is the problem related to shortest paths?

• Yes: The objective is to minimize the total price for a specified number of trips, which translates into finding least-cost paths within a graph structure.

Is the graph weighted?

• Yes: Each flight has a different cost, hence the graph is weighted.

Does the problem have small constraints?

• Yes: The description and nature of the question suggest a manageable number of cities and flights, making exhaustive methods like DFS feasible.

Conclusion: Although initially, a path finding algorithm like Dijkstra's might seem appropriate due to the weighted graph, the small constraints hint that a more exhaustive or state-encompassing method like Depth-First Search (DFS) or backtracking could be effectively utilized, especially for complex path-finding with additional constraints, as indicated by our flowchart analysis. This assumes the problem complexity and constraints are small enough that such an approach would remain computationally reasonable.

Intuition

The intuition behind the solution lies in understanding that we have two key tasks:

1. Track the paths taken during the trips to see how many times each node is visited.
2. Decide which nodes' prices should be halved to minimize the overall cost, taking into account their frequency of being on a trip path and ensuring the chosen nodes are not adjacent.

To solve the first task, we use a Depth-First Search (DFS) approach. We perform DFS for each trip to mark the frequency at which each node is visited on any path from the start to the end of that trip.

For the second task, we again apply a DFS approach. This time, we calculate for each node the total price of the sub-tree rooted at that node with and without halving its price. To ensure that prices are halved only for non-adjacent nodes, we compare the total cost with halved price at each node with the sum of minimum costs from its children.

Overall, by using DFS, we can ensure we visit each node and process them while adhering to the given constraints of the problem.

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

Solution Approach

The solution uses depth-first search (DFS), a graph traversal algorithm, to explore the tree and solve the two main tasks: tracking paths and minimizing the total price sum.

Here are the steps that comprise the solution approach:

1. Building the Graph: The graph is represented using an adjacency list, which is a common efficient way to store a graph when dealing with sparse trees. Each node is indexed from 0 to n - 1, and the adjacency list g maintains a list of children for every node.

2. Counting Visits (DFS): A DFS is performed through the function dfs(i: int, fa: int, k: int). This function is called for each trip to track how many times the nodes are visited across all trips. The function uses two parameters, i for the current node and fa for the parent node, to avoid revisiting the parent node during the recursion. The count for each node is increased when it is visited, and if we reach the end node of the trip (k), we return True. If the current path does not lead to k, we undo the increment on cnt[i] before backtracking. This process is performed for all trips and effectively maps the usage frequency of each node across all trips.

3. Halving Prices and Cost Calculation (DFS): Another DFS function, dfs2(i: int, fa: int), is called to find the sum of prices for the sub-tree rooted at each node i, considering the option to halve the node's price. The function calculates two costs: a, which is the cost without any price halving, and b, which is with the node's price halved. For each child j, we recursively call dfs2(j, i) to obtain the optimal costs for the child's sub-tree and accumulate these costs. The crucial part is to choose the minimum between x and y for the cost a to reflect the best decision for non-adjacent nodes' price reduction, while b always includes the lower child costs as we're halving i's price.

4. Final Minimum Cost: The final step is to call dfs2(0, -1) for the root of the tree (noting that -1 indicates there is no parent for the root) to get the minimum overall price with and without halving the price of the root node. The minimum of these two values is the answer to the problem as we track the minimum cost for the whole tree.

The use of Python's Counter from the collections module helps in handling the node visit counts with convenience, avoiding manual initialization and incrementing of counts for every node. The recursive nature of DFS elegantly handles the complex conditions of the problem by breaking it down into smaller sub-problems of the tree.

Example Walkthrough

Let's illustrate the solution approach with a small example:

Suppose we have a tree with 5 nodes where the prices are as follows: [2, 1, 3, 2, 2] and the edges list that defines the connections is [[0, 1], [0, 2], [2, 3], [2, 4]]. We have 2 trips: one from node 1 to 3 and another from node 1 to 4.

The adjacency list g for our tree would look like this:

{
0: [1, 2],
1: [0],
2: [0, 3, 4],
3: [2],
4: [2]
}

1. Building the Graph: We create the adjacency list g to represent our tree based on the edges.

2. Counting Visits (DFS): We perform DFS to count visits for each trip:

   • For trip 1 to 3, DFS traversal is as follows: 1 -> 0 -> 2 -> 3. We increment the visit count for these nodes.
   • For trip 1 to 4, DFS traversal is: 1 -> 0 -> 2 -> 4. We increment the visit count for these nodes.

After counting visits, our visit counts (ignoring nodes with 0 visits) might look like: {1: 2, 0: 2, 2: 2, 3: 1, 4: 1}.

3. Halving Prices and Cost Calculation (DFS): We perform another DFS to calculate the optimal cost for each node's sub-tree:
   • Starting with dfs2(0, -1), we consider halving the price of node 0. We find the cost of sub-trees 1, 3, and 4 through recursive calls.
   • For node 2, we recursively find the best cost of sub-trees 3 and 4, and so on for each node.

By the end of this step, we have determined the minimum cost for each node's sub-tree, considering the option of halving the node's price.

4. Final Minimum Cost: The result of dfs2(0, -1) would give us two values: the minimum cost of trips without halving the price of the root, and the minimum cost of trips with the root price halved. The minimum of these two values will give us the final answer to the problem.

By following these steps and applying DFS effectively, we track each node’s frequency of visit and decide strategically which non-adjacent nodes should have their prices halved to minimize the total cost of all trips. In this example, the result might indicate halving the price of either the most frequently visited non-adjacent node or the node that gives the maximum saving when its price is halved, depending on the specific visit counts and node prices.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code consists of two parts: building the graph and performing depth-first search (DFS) operations.

• Building the graph from the input edges requires iterating through each edge once. Since there are E edges provided where E is the length of the edges list, this part has a complexity of O(E).

• The dfs function is called for each trip to find the path from the start to the end. The path for a given trip is found by traversing the graph using DFS. In the worst case, this can go up to O(n) for each trip because we need to visit all the nodes. Since there are T trips provided where T is the length of the trips list, this part has a complexity of O(T * n).

• The dfs2 function is called once and it potentially visits each node once and computes the minimum price for each node. This part has a complexity of O(n) since it runs a DFS on the graph with n nodes.

Combining these, the total time complexity is O(E + T * n + n). Since E is typically less than n^2 and T could be up to n, the combined complexity could be viewed as O(n^2) in the worst case when considering dense graphs and a high number of trips.

Space Complexity

The space complexity is dominated by the space needed to store the graph, the cnt counter, and the recursive stack for the DFS operations.

• The graph is represented as an adjacency list, which in the worst case stores an entry for each edge twice (since the graph is undirected). Therefore, the space complexity for the graph is O(E).

• The cnt Counter object keeps a count of nodes for the number of trips that pass through them. In the worst case, all trips will involve all nodes, thus the Counter would have n entries. This would take O(n) space.

• The recursive stack for the dfs and dfs2 functions would at worst case store a context for each node in a chain-like sequence of nodes (O(n) space).

The combined space complexity hence is O(E + n + n) which simplifies to O(E + n). Since E can at most be n(n-1)/2 for a complete graph, the space complexity can be O(n^2) in the worst case.

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