Cheapest Flights Within K Stops
Recognize the pattern
Brute force idea
A straightforward first read of Cheapest Flights Within K Stops is this: Try all paths with ≤ k+1 edges — exponential. Enumerates every possible route. That instinct is useful because it follows the prompt literally, but it usually keeps revisiting work the problem is begging you to organize.
Better approach
A calmer way to see Cheapest Flights Within K Stops is this: Bellman-Ford with k+1 relaxation rounds. Or BFS with (cost, node, stops) states and a min-heap (modified Dijkstra). Each round relaxes all edges, but only for k+1 rounds. The goal is not to be clever for its own sake, but to remember the one relationship that keeps the solution grounded as you move forward.
Key invariant
The truth you want to protect throughout Cheapest Flights Within K Stops is this: After i rounds of Bellman-Ford, the shortest paths using at most i edges are computed. Running k+1 rounds gives the answer with the stop constraint. If that remains true after every update, the rest of the reasoning has a stable place to stand.
Watch out for
The trap in Cheapest Flights Within K Stops usually looks like this: Using standard Dijkstra without the stop constraint — Dijkstra finds the cheapest path regardless of stops. You need to track stops as part of the state. When the code becomes mechanical before the idea is clear, small edge cases start breaking the whole story.