What is the optimal time complexity of solving Cheapest Flights Within K Stops?

The optimized solution achieves a time complexity of O(n) by using sorting, mathematical shortcuts, or hash table lookups.

What is the auxiliary space complexity for Cheapest Flights Within K Stops?

The space complexity is O(1) since we only use a minimal/constant amount of extra memory for tracking variables (or auxiliary space if dynamic structures are allocated).

What are the typical edge cases we must handle in Cheapest Flights Within K Stops?

Key edge cases include empty collections, negative or large bounds causing integer overflow, arrays with only one element, or inputs where target criteria are not met.

How can we optimize the brute force approach for Cheapest Flights Within K Stops?

Brute force methods often inspect all possible combinations. We optimize this by sorting the input list to enable binary search, or tracking processed items in a set to avoid redundant nested loop lookups.

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Cheapest Flights Within K Stops

Learn how to solve the 'Cheapest Flights Within K Stops' problem. This detailed resource details brute force and optimized approaches.

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Problem Statement

Easy

There are n cities connected by some number of flights. You are given an array flights where flights[i] = [from_i, to_i, price_i] indicates that there is a flight from city from_i to city to_i with cost price_i.

You are also given three integers src, dst, and k, return the cheapest price from src to dst with at most k stops. If there is no such route, return -1.

Write a function findCheapestPrice(n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int.

Constraints

•1 <= n <= 100
•0 <= len(flights) <= (n * (n - 1) / 2)
•flights[i].length == 3
•0 <= from_i, to_i < n
•1 <= price_i <= 10^4
•0 <= src, dst < n
•src != dst
•0 <= k < n

Examples

Example 1

Input

n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1

Output

Explanation

The cheapest path from city 0 to city 3 with at most 1 stop is 0 -> 1 -> 3 with cost 100 + 600 = 700.

Example 2

Input

n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1

Output

Explanation

Cheapest path is 0 -> 1 -> 2 with cost 200, which has exactly 1 stop.

Need a Hint?

Represent graph node connections as an adjacency list/matrix, then use standard BFS or DFS graph traversal.

Edge Cases to Watch

Empty list or null input variables
Single item lists/arrays
Extremely large input bounds causing integer or stack overflow

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