How do you solve Minimum Cost to Reach Destination in Time on LeetCode?

Minimum Cost to Reach Destination in Time (LeetCode #1928) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(E log V) time complexity and O(V) space complexity. Key insight: Using a priority queue allows us to efficiently explore the most promising paths first.

What is the brute force solution for Minimum Cost to Reach Destination in Time?

The brute force approach for Minimum Cost to Reach Destination in Time is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach explores all possible paths from city 0 to city n-1, calculating the total travel time and cost for each path. This method guarantees finding the correct answer but is inefficient for larger graphs due to its exhaustive nature. The optimal Optimal Solution approach improves this to O(E log V).

What algorithm or data structure is used to solve Minimum Cost to Reach Destination in Time?

Minimum Cost to Reach Destination in Time uses the following concepts: Array, Dynamic Programming, Graph Theory. The recommended patterns to study are: Graph Traversal, Priority Queue, Dynamic Programming.

What are the interview tips for Minimum Cost to Reach Destination in Time?

Always clarify the constraints and edge cases before diving into the solution. Explain your thought process and why you choose a particular approach. Practice implementing Dijkstra's algorithm and understand its variations.

What are common mistakes when solving Minimum Cost to Reach Destination in Time?

Failing to account for the time constraint when exploring paths. Not using a priority queue effectively, leading to suboptimal performance.

#1928

Minimum Cost to Reach Destination in Time

Hard

ArrayDynamic ProgrammingGraph TheoryGraph TraversalPriority QueueDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(E log V)
Space	O(1)	O(V)

💡

Intuition

Time O(E log V)Space O(V)

The optimal approach uses a modified Dijkstra's algorithm to efficiently find the minimum cost path while considering both travel time and passing fees. This method leverages a priority queue to explore the most promising paths first, ensuring we find the minimum cost without exhaustively searching every path.

⚙️

Algorithm

4 steps

1Step 1: Create a graph representation from the edges.
2Step 2: Use a priority queue to explore paths, storing the current city, total time, and total cost.
3Step 3: For each city, if the current time is within maxTime, update the cost and explore neighboring cities.
4Step 4: Continue until reaching city n-1 or exhausting all options, returning the minimum cost found.

solution.py24 lines

1import heapq
2from collections import defaultdict
3
4def minCost(maxTime, edges, passingFees):
5    graph = defaultdict(list)
6    for x, y, time in edges:
7        graph[x].append((y, time))
8        graph[y].append((x, time))
9
10    pq = [(passingFees[0], 0, 0)]  # (cost, time, city)
11    min_cost = {(0, 0): passingFees[0]}
12
13    while pq:
14        cost, time, city = heapq.heappop(pq)
15        if city == len(passingFees) - 1:
16            return cost
17        for neighbor, travel_time in graph[city]:
18            new_time = time + travel_time
19            new_cost = cost + passingFees[neighbor]
20            if new_time <= maxTime:
21                if (neighbor, new_time) not in min_cost or new_cost < min_cost[(neighbor, new_time)]:
22                    min_cost[(neighbor, new_time)] = new_cost
23                    heapq.heappush(pq, (new_cost, new_time, neighbor))
24    return -1

ℹ

Complexity note: The time complexity is O(E log V) due to the priority queue operations, where E is the number of edges and V is the number of vertices (cities). The space complexity is O(V) for storing the graph and the minimum costs.

1Using a priority queue allows us to efficiently explore the most promising paths first.
2Combining travel time and passing fees into a single cost metric helps streamline the search process.

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