# Cheapest Flights Within K Stops
There are `n` cities connected by flights `[from, to, price]`. Given a source, destination, and a maximum of `k` stops, return the cheapest price to get from `src` to `dst` within `k` stops (or `-1` if impossible.)
A natural escalation past Mock 4's plain Network Delay Time (Dijkstra): **plain Dijkstra does not correctly handle a hop-count constraint**, because it always finalizes the globally cheapest path first, which may use more stops than allowed. Use a Bellman-Ford-style relaxation instead, capped at `k + 1` rounds.
**Example:** `n=3, flights=[[0,1,100],[1,2,100],[0,2,500]], src=0, dst=2, k=1` -> `200`.
Target complexity: O(K * E).