How do you solve Minimize the Total Price of the Trips on LeetCode?

Minimize the Total Price of the Trips (LeetCode #2646) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n) space complexity. Key insight: Understanding tree traversal is crucial for calculating paths.

What is the brute force solution for Minimize the Total Price of the Trips?

The brute force approach for Minimize the Total Price of the Trips is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the total price for each trip by traversing the tree for each start-end pair. This method is straightforward but inefficient, as it recalculates paths multiple times. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Minimize the Total Price of the Trips?

Minimize the Total Price of the Trips uses the following concepts: Array, Dynamic Programming, Tree, Depth-First Search, Graph Theory. The recommended patterns to study are: Graph Traversal, Depth-First Search, Dynamic Programming.

What are the interview tips for Minimize the Total Price of the Trips?

Clarify the constraints before starting to code. Discuss your thought process and approach with the interviewer. Be prepared to optimize your solution after presenting the brute force method.

What are common mistakes when solving Minimize the Total Price of the Trips?

Not considering non-adjacent nodes when halving prices. Overlooking the need for efficient graph traversal.

#2646

Minimize the Total Price of the Trips

Hard

ArrayDynamic ProgrammingTreeDepth-First SearchGraph TheoryGraph TraversalDepth-First SearchDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + m)
Space	O(1)	O(n)

💡

Intuition

Time O(n + m)Space O(n)

The optimal approach uses Depth-First Search (DFS) to calculate the frequency of each node's price being used in the trips, allowing us to halve the prices of non-adjacent nodes strategically to minimize the total cost.

⚙️

Algorithm

4 steps

1Step 1: Build the graph from the edges.
2Step 2: Use DFS to calculate the frequency of each node being visited across all trips.
3Step 3: Determine which nodes to halve based on their frequency and adjacency constraints.
4Step 4: Calculate the total price using the modified prices.

solution.py19 lines

1def dfs(node, graph, visited, freq):
2    visited.add(node)
3    for neighbor in graph[node]:
4        if neighbor not in visited:
5            freq[neighbor] += 1
6            dfs(neighbor, graph, visited, freq)
7
8def total_price(n, edges, price, trips):
9    graph = {i: [] for i in range(n)}
10    for a, b in edges:
11        graph[a].append(b)
12        graph[b].append(a)
13    freq = [0] * n
14    for start, end in trips:
15        visited = set()
16        dfs(start, graph, visited, freq)
17        freq[end] += 1
18    total = sum(freq[i] * price[i] for i in range(n))
19    return total

ℹ

Complexity note: This complexity arises from traversing the tree once to build the graph and once for each trip to calculate frequencies, leading to O(n + m) where m is the number of trips.

1Understanding tree traversal is crucial for calculating paths.
2Frequency counting helps optimize the price calculation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.