2646. Minimize the Total Price of the Trips
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:
- Track the paths taken during the trips to see how many times each node is visited.
- 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:
-
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
ton - 1
, and the adjacency listg
maintains a list of children for every node. -
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 andfa
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 returnTrue
. If the current path does not lead tok
, we undo the increment oncnt[i]
before backtracking. This process is performed for all trips and effectively maps the usage frequency of each node across all trips. -
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 nodei
, considering the option to halve the node's price. The function calculates two costs:a
, which is the cost without any price halving, andb
, which is with the node's price halved. For each childj
, we recursively calldfs2(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 betweenx
andy
for the costa
to reflect the best decision for non-adjacent nodes' price reduction, whileb
always includes the lower child costs as we're halvingi
's price. -
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.
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Start EvaluatorExample 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:
1{ 20: [1, 2], 31: [0], 42: [0, 3, 4], 53: [2], 64: [2] 7}
-
Building the Graph: We create the adjacency list
g
to represent our tree based on theedges
. -
Counting Visits (DFS): We perform DFS to count visits for each trip:
- For trip
1
to3
, DFS traversal is as follows:1 -> 0 -> 2 -> 3
. We increment the visit count for these nodes. - For trip
1
to4
, DFS traversal is:1 -> 0 -> 2 -> 4
. We increment the visit count for these nodes.
- For trip
After counting visits, our visit counts (ignoring nodes with 0 visits) might look like: {1: 2, 0: 2, 2: 2, 3: 1, 4: 1}
.
- 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 node0
. We find the cost of sub-trees1
,3
, and4
through recursive calls. - For node
2
, we recursively find the best cost of sub-trees3
and4
, and so on for each node.
- Starting with
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.
- 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
1from typing import List
2from collections import Counter
3
4class Solution:
5 def minimumTotalPrice(self, n: int, edges: List[List[int]], price: List[int], trips: List[List[int]]) -> int:
6 # A DFS helper function to count the number of trips passing through each node.
7 def dfs(current_node: int, parent_node: int, destination: int) -> bool:
8 trip_count[current_node] += 1
9 if current_node == destination:
10 return True
11
12 path_exists = any(
13 neighbor != parent_node and dfs(neighbor, current_node, destination)
14 for neighbor in graph[current_node]
15 )
16
17 if not path_exists:
18 trip_count[current_node] -= 1
19
20 return path_exists
21
22 # A DFS helper function to compute the minimum total price to visit all nodes,
23 # considering the count of trips passing through each node.
24 def dfs_min_price(current_node: int, parent_node: int) -> (int, int):
25 full_price = trip_count[current_node] * price[current_node]
26 half_price = full_price // 2
27
28 for neighbor in graph[current_node]:
29 if neighbor != parent_node:
30 min_full, min_half = dfs_min_price(neighbor, current_node)
31 full_price += min(min_full, min_half)
32 half_price += min_full
33
34 return full_price, half_price
35
36 # Construct the graph from edges.
37 graph = [[] for _ in range(n)]
38 for a, b in edges:
39 graph[a].append(b)
40 graph[b].append(a)
41
42 # Count the number of trips that pass through each node.
43 trip_count = Counter()
44 for start, end in trips:
45 dfs(start, -1, end)
46
47 # Calculate the minimum total price considering both full price and half price tickets.
48 full_price, half_price = dfs_min_price(0, -1)
49 return min(full_price, half_price)
50
51
52# Note: The comments have been added to explain the functions and logic used in the code.
53# All variable and method names are kept consistent with Python 3 standards.
54
1class Solution {
2 private List<Integer>[] graph;
3 private int[] price;
4 private int[] count;
5
6 public int minimumTotalPrice(int n, int[][] edges, int[] price, int[][] trips) {
7 this.price = price;
8 count = new int[n];
9 graph = new List[n];
10 Arrays.setAll(graph, k -> new ArrayList<>());
11 // Construct the undirected graph
12 for (int[] edge : edges) {
13 int from = edge[0], to = edge[1];
14 graph[from].add(to);
15 graph[to].add(from);
16 }
17 // Count the frequencies of nodes being on the path from start to end for each trip
18 for (int[] trip : trips) {
19 int start = trip[0], end = trip[1];
20 depthFirstSearch(start, -1, end);
21 }
22 // Calculate the minimum price
23 int[] answer = secondDepthFirstSearch(0, -1);
24 return Math.min(answer[0], answer[1]);
25 }
26
27 // Helper DFS method to count the frequency of each node on the path from 'start' to 'k'
28 private boolean depthFirstSearch(int current, int parent, int destination) {
29 ++count[current];
30 if (current == destination) {
31 return true;
32 }
33 boolean found = false;
34 for (int neighbor : graph[current]) {
35 if (neighbor != parent) {
36 found = depthFirstSearch(neighbor, current, destination);
37 if (found) {
38 break;
39 }
40 }
41 }
42 if (!found) {
43 --count[current];
44 }
45 return found;
46 }
47
48 // Helper DFS method to compute two prices from each node to child nodes:
49 // 1. The price of traveling through node 'i' based on the frequency.
50 // 2. The price of travel avoiding node 'i' if possible.
51 private int[] secondDepthFirstSearch(int current, int parent) {
52 int directTotalPrice = count[current] * price[current]; // price if using this node
53 int bypassedTotalPrice = directTotalPrice >> 1; // half price if bypassing this node
54 for (int neighbor : graph[current]) {
55 if (neighbor != parent) {
56 int[] prices = secondDepthFirstSearch(neighbor, current);
57 directTotalPrice += Math.min(prices[0], prices[1]); // min price from this node to its children
58 bypassedTotalPrice += prices[0]; // full price if this node is bypassed
59 }
60 }
61 return new int[] {directTotalPrice, bypassedTotalPrice};
62 }
63}
64
1#include<vector>
2#include<functional>
3using namespace std;
4
5class Solution {
6public:
7 // Function to find the minimum total price to pay for the trips.
8 int minimumTotalPrice(int n, vector<vector<int>>& edges, vector<int>& price, vector<vector<int>>& trips) {
9 // Adjacency list to represent the graph.
10 vector<vector<int>> graph(n);
11 // Count of how many times a node is visited during the trips.
12 vector<int> visitCount(n);
13
14 // Build the undirected graph from the edges.
15 for (auto& edge : edges) {
16 int a = edge[0], b = edge[1];
17 graph[a].push_back(b);
18 graph[b].push_back(a);
19 }
20
21 // Depth-first search to increment the visit count on the path from start to end.
22 // i: current node, parent: parent node, destination: destination node.
23 function<bool(int, int, int)> dfs = [&](int currentNode, int parentNode, int destination) -> bool {
24 visitCount[currentNode]++;
25 if (currentNode == destination) {
26 return true;
27 }
28 bool pathExists = false;
29 for (int neighbor : graph[currentNode]) {
30 if (neighbor != parentNode) {
31 pathExists = dfs(neighbor, currentNode, destination);
32 if (pathExists) {
33 break; // If the destination is found, break out of the loop.
34 }
35 }
36 }
37 if (!pathExists) {
38 visitCount[currentNode]--; // If no path, decrement the visit count.
39 }
40 return pathExists;
41 };
42
43 // Second depth-first search to compute the total price to pay.
44 // i: current node, parent: parent node.
45 function<pair<int, int>(int, int)> dfs2 = [&](int currentNode, int parentNode) -> pair<int, int> {
46 int fullPrice = visitCount[currentNode] * price[currentNode];
47 int halfPrice = fullPrice >> 1; // Bitwise shift to divide by 2.
48 for (int neighbor : graph[currentNode]) {
49 if (neighbor != parentNode) {
50 auto [fullNeighborPrice, halfNeighborPrice] = dfs2(neighbor, currentNode);
51 fullPrice += min(fullNeighborPrice, halfNeighborPrice);
52 halfPrice += fullNeighborPrice;
53 }
54 }
55 return {fullPrice, halfPrice};
56 };
57
58 // Run the first depth-first search for each trip to update the visit count.
59 for (auto& trip : trips) {
60 int start = trip[0], end = trip[1];
61 dfs(start, -1, end);
62 }
63
64 // Call the second depth-first search from the root node to calculate the total price.
65 auto [fullPriceResult, halfPriceResult] = dfs2(0, -1);
66
67 // Return the minimum of the two calculated prices.
68 return min(fullPriceResult, halfPriceResult);
69 }
70};
71
1// Function to calculate the minimum total price for all trips.
2// n: number of locations, edges: list of connections between locations,
3// price: array where each index represents a price for the location,
4// trips: list of trips represented by [start, end],
5// returns the minimum total price for all trips.
6function minimumTotalPrice(
7 n: number,
8 edges: number[][],
9 price: number[],
10 trips: number[][],
11): number {
12 // Create a graph as an adjacency list.
13 const graph: number[][] = Array.from({ length: n }, () => []);
14 for (const [from, to] of edges) {
15 graph[from].push(to);
16 graph[to].push(from);
17 }
18
19 // Initialize an array to count the number of trips through each location.
20 const tripCount: number[] = new Array(n).fill(0);
21
22 // Depth-first search function to count trip occurrences between start and end.
23 // i: current location, parent: parent location, destination: trip's destination:
24 // returns boolean indicating if the destination was reached in the search.
25 const depthFirstSearch = (i: number, parent: number, destination: number): boolean => {
26 tripCount[i]++;
27 if (i === destination) {
28 return true;
29 }
30 let reachedDestination = false;
31 for (const next of graph[i]) {
32 if (next !== parent) {
33 reachedDestination = depthFirstSearch(next, i, destination);
34 if (reachedDestination) {
35 break;
36 }
37 }
38 }
39 if (!reachedDestination) {
40 tripCount[i]--;
41 }
42 return reachedDestination;
43 };
44
45 // Count trip occurrences for each trip in the trips array.
46 for (const [start, end] of trips) {
47 depthFirstSearch(start, -1, end);
48 }
49
50 // Second depth-first search function to calculate the total cost with discounts.
51 // i: current location, parent: parent location:
52 // returns array with total cost without discount, and with discount.
53 const calculateCosts = (i: number, parent: number): number[] => {
54 let totalCostWithoutDiscount: number = price[i] * tripCount[i];
55 let totalCostWithDiscount: number = totalCostWithoutDiscount >> 1;
56 for (const next of graph[i]) {
57 if (next !== parent) {
58 const [costWithoutDiscount, costWithDiscount] = calculateCosts(next, i);
59 totalCostWithoutDiscount += Math.min(costWithoutDiscount, costWithDiscount);
60 totalCostWithDiscount += costWithoutDiscount;
61 }
62 }
63 return [totalCostWithoutDiscount, totalCostWithDiscount];
64 };
65
66 // Calculate the costs from the starting location.
67 const [costWithoutDiscount, costWithDiscount] = calculateCosts(0, -1);
68
69 // Return the minimum of the two calculated costs.
70 return Math.min(costWithoutDiscount, costWithDiscount);
71}
72
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 whereE
is the length of the edges list, this part has a complexity ofO(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 toO(n)
for each trip because we need to visit all the nodes. Since there areT
trips provided whereT
is the length of the trips list, this part has a complexity ofO(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 ofO(n)
since it runs a DFS on the graph withn
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 haven
entries. This would takeO(n)
space. -
The recursive stack for the
dfs
anddfs2
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.
What data structure does Breadth-first search typically uses to store intermediate states?
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