Cheapest Flights Within K Stops

def findCheapestPrice(n, flights, src, dst, k):
    out = [[] for _ in range(n)]
    for a, b, price in flights:
        out[a].append((b, price))
    best = -1
    # Each entry is a route in progress: (city, price so far, flights taken).
    stack = [(src, 0, 0)]
    while stack:
        city, cost, used = stack.pop()
        if city == dst:
            if best == -1 or cost < best:
                best = cost
            continue
        # k stops allow at most k+1 flights.
        if used == k + 1:
            continue
        for nxt, price in out[city]:
            stack.append((nxt, cost + price, used + 1))
    return best

def findCheapestPrice(n, flights, src, dst, k):
    INF = float("inf")
    # cost[c]: the cheapest price to reach city c with the flights allowed so far.
    cost = [INF] * n
    cost[src] = 0
    # Each round allows one more flight; k stops means k+1 flights.
    for _ in range(k + 1):
        # Read last round's prices and write into a copy,
        # so one round can never chain two flights.
        nxt = cost[:]
        changed = False
        for a, b, price in flights:
            if cost[a] != INF and cost[a] + price < nxt[b]:
                nxt[b] = cost[a] + price
                changed = True
        cost = nxt
        if not changed:
            break
    return -1 if cost[dst] == INF else cost[dst]