Cheapest Flights Within K Stops

Write `find_cheapest_price(n, flights, src, dst, k)` returning the cheapest price from `src` to `dst` using at most `k` stops, or `-1`. n=3, flights=[[0,1,100],[1,2,100],[0,2,500]], src=0, dst=2, k=0 -> 500 Dijkstra finalises a node at its cheapest cost, which may use too many stops — and a pricier route with fewer stops can be the only legal one. Bellman-Ford relaxes all edges `k + 1` times, so the hop bound is built into the iteration count. Relax from a *snapshot* of the previous round, or one round can chain several hops.