LeetCode 787 Cheapest Flights Within K Stops - Algorithm

$1.$ Initialize dist[src] = 0, all others to infinity.

$2.$ Repeat $k+1$ times: copy $dist$ to prev, then for each edge $(u, v, w)$, if prev[u] + w < dist[v], set dist[v] = prev[u] + w.

$3.$ After $k+1$ iterations, return dist[dst] if finite, otherwise return $-1$.

The two-array approach ensures you count stops correctly. Each iteration represents one additional flight (edge) in the path.