The O(n) Club: Cheapest Flights Within K Stops: Layovers, Logic, and Code Turbulence

The O(n) Club: Cheapest Flights Within K Stops – Layovers, Logic, and Code Turbulence

⚡ TL;DR

Want the cheapest ticket from A to B with no more than K layovers? Try every possible flight combo if you hate free time or flex your Bellman-Ford muscles right. Here’s a sneak peek (Java, naturally):

// Brute-ish but gets the point:
int findCheapestPrice(int n, int[][] flights, int src, int dst, int K) {
    int[] cost = new int[n];
    Arrays.fill(cost, Integer.MAX_VALUE);
    cost[src] = 0;
    for (int i = 0; i <= K; i++) {
        int[] temp = Arrays.copyOf(cost, n);
        for (int[] f : flights) {
            int u = f[0], v = f[1], w = f[2];
            if (cost[u] != Integer.MAX_VALUE)
                temp[v] = Math.min(temp[v], cost[u] + w);
        }
        cost = temp;
    }
    return cost[dst] == Integer.MAX_VALUE ? -1 : cost[dst];
}

🧠 How Devs Usually Mess This Up

Welcome to the league of Almost Right™:

🔍 What’s Actually Going On

Flight search, but with rules. Picture a world map where airlines charge like road tolls, and you only get K bathroom breaks (aka layovers) on your trip—or Mom gets mad. Bellman-Ford runs an annual Sale: Every ‘iteration’ is a shot to add another flight to your itinerary, but only if it fits the layover budget.
Think of the cost array like your itinerary list: use yesterday’s version for each update, or you’ll sneak in extra stops and get mugged by the algorithmic TSA.

🛠️ PseudoCode

• Start with all flights overpriced (∞), except your home airport (0):
  Because staying home costs nothing unless you count dignity.
• Do K+1 rounds (flights):
  • Make a backup of current least-cost array.
  • For every flight u → v costing w:
  • If you could get to u in ≤ current cost, see if v is cheaper via u + w.
  • Only update the backup! No time travel allowed.
  • After the round, backup is the new base. Lather, rinse, repeat.
• Endgame: If you ever reach your destination, congrats! Otherwise, consider carrier pigeons.
• Java flavor:

  // See TL;DR above. Highlight: Bellman-Ford, but capped at K+1 rounds.

💻 The Code

import java.util.*;
 public class CheapestFlights {
    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int k) {
        int[] cost = new int[n];
        Arrays.fill(cost, Integer.MAX_VALUE);
        cost[src] = 0;
        for (int i = 0; i <= k; i++) {
            int[] next = Arrays.copyOf(cost, n);
            for (int[] f : flights) {
                int u = f[0], v = f[1], w = f[2];
                if (cost[u] != Integer.MAX_VALUE && cost[u] + w < next[v]) {
                    next[v] = cost[u] + w;
                }
            }
            cost = next;
        }
        return cost[dst] == Integer.MAX_VALUE ? -1 : cost[dst];
    }
     public static void main(String[] args) {
        CheapestFlights cf = new CheapestFlights();
        int[][] flights = { {0,1,100}, {1,2,100}, {0,2,500} };
        System.out.println(cf.findCheapestPrice(3, flights, 0, 2, 1)); // 200
        System.out.println(cf.findCheapestPrice(3, flights, 0, 2, 0)); // 500
    }
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Negative flight cost? Bellman-Ford isn’t fazed, but your travel agent might be.
• No solution? Return -1 (no, hiking isn’t supported).
• K = 0? Only nonstop flights or bust. Hope you packed snacks.
• Forgetting to copy cost arrays: Extra stops sneak in. Interviewers sigh.
• Time/space: O(K*E) time, O(N) space. Like waiting in airport security, but you know it ends.

🧠 Interviewer Brain-Tattoo

“If you forget to back up your costs during Bellman-Ford, your layovers will multiply faster than airline fees.”