Cheapest Flights Within K Stops - Visualization

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ProblemExamplesSolutionAlgorithmCode

Problem Statement❯

You are given information about n cities connected by m flights. Each flight is represented as [fromi, toi, pricei] indicating a direct flight from city fromi to city toi with a cost of pricei.

You are also given three integers:

  • src – the starting city
  • dst – the target city
  • k – the maximum number of allowed stops (i.e., you can take at most k + 1 flights)

Your goal is to find the cheapest price to travel from src to dst with at most k stops. If such a route doesn't exist, return -1.

Examples❯

n Flights src dst k Cheapest Price
4 [[0,1,100],[1,2,100],[2,3,100],[0,3,500]] 0 3 1 500
4 [[0,1,100],[1,2,100],[2,3,100],[0,3,500]] 0 3 2 300
3 [[0,1,200],[1,2,200],[0,2,500]] 0 2 0 500
3 [[0,1,100],[1,2,100]] 0 2 0 -1

Solution❯

Understanding the Problem

We are given a set of flights between cities, each with a certain cost. Our goal is to find the cheapest flight route from a source city to a destination city, but with a twist — we are only allowed to make at most K stops along the way (that is, up to K + 1 cities including source and destination).

This is a variation of the classic shortest path problem, but with a constraint on the number of edges (stops) we can traverse. We model the cities as nodes in a graph, and flights as directed, weighted edges.

Step-by-Step Solution with Example

step 1: Represent the graph

We convert the list of flights into an adjacency list where each key is a source city and the value is a list of destination cities with their respective flight costs.


flights = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1

graph = {
  0: [(1, 100), (2, 500)],
  1: [(2, 100)]
}

step 2: Choose a strategy to explore the graph

Since we are constrained by the number of stops, traditional Dijkstra’s algorithm won’t directly work. Instead, we use a modified Breadth-First Search (BFS) that explores paths while also tracking the number of stops and total cost.

step 3: Use a queue to perform BFS

Each queue entry contains:

  • Current city
  • Total cost to reach this city
  • Number of stops made so far
We initialize the queue with (0, 0, 0) (starting city, cost, stops).

step 4: Track the minimum cost to reach each city with stops

We use a dictionary like minCost[(city, stops)] = cost to ensure we don’t revisit worse paths.

step 5: Explore the graph

While the queue is not empty:

  1. Pop the front of the queue
  2. If it's the destination, track the cost
  3. If stops exceed k, skip
  4. For each neighbor, if going there is cheaper or not yet visited with this stop count, enqueue it

step 6: Apply this to the example

Start at city 0 with 0 cost and 0 stops:

  • From 0 → 1 (cost 100), stops = 1 → Enqueue (1, 100, 1)
  • From 0 → 2 (cost 500), stops = 1 → Enqueue (2, 500, 1)
Now at (1, 100, 1):
  • From 1 → 2 (cost 100 more), total = 200, stops = 2 → Enqueue (2, 200, 2)
(2, 200, 2) is a valid answer because stops ≤ k + 1. Answer: 200

Edge Cases

  • No valid route within K stops: If all paths exceed the allowed stops, we return -1
  • Zero stops allowed (K=0): Only direct flights are considered
  • Multiple flights between same cities: Always pick the cheaper one while processing neighbors
  • Disconnected cities: Ensure unreachable nodes are handled correctly

Finally

This problem beautifully demonstrates how classic graph traversal algorithms can be adapted to incorporate real-world constraints like number of stops. By blending BFS with cost and stop tracking, we efficiently explore valid routes and select the cheapest one. Always break down the problem into steps — understand the data, model the graph, and use the right traversal strategy.

Algorithm Steps❯

  1. Build an adjacency list to represent the graph of cities and flights.
  2. Initialize a queue with a tuple (src, 0, 0) representing (current city, cost so far, stops so far).
  3. Maintain a minCost map to record the minimum cost to reach a node with a certain number of stops.
  4. While the queue is not empty:
    1. Dequeue a tuple (city, cost, stops).
    2. If city == dst, update the result if cost is lower.
    3. If stops > k, continue to next iteration.
    4. For each neighbor of the current city, push (neighbor, cost + price, stops + 1) to the queue if this new cost is cheaper than any previously recorded cost to that neighbor with same or fewer stops.
  5. If a path is found, return the cheapest cost; otherwise, return -1.

Code

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#include <stdio.h>
#include <limits.h>

#define MAXN 100
#define MAXK 100

int min(int a, int b) {
    return a < b ? a : b;
}

int findCheapestPrice(int n, int flights[][3], int flightsSize, int src, int dst, int k) {
    int dist[MAXN][MAXK + 2];
    for (int i = 0; i < n; i++)
        for (int j = 0; j <= k + 1; j++)
            dist[i][j] = INT_MAX;
    dist[src][0] = 0;

    for (int i = 0; i <= k; i++) {
        for (int j = 0; j < flightsSize; j++) {
            int u = flights[j][0], v = flights[j][1], w = flights[j][2];
            if (dist[u][i] != INT_MAX)
                dist[v][i + 1] = min(dist[v][i + 1], dist[u][i] + w);
        }
    }

    int ans = INT_MAX;
    for (int i = 1; i <= k + 1; i++)
        ans = min(ans, dist[dst][i]);
    return ans == INT_MAX ? -1 : ans;
}

int main() {
    int flights[][3] = {{0,1,100},{1,2,100},{2,3,100},{0,3,500}};
    printf("%d\n", findCheapestPrice(4, flights, 4, 0, 3, 1)); // 500
    printf("%d\n", findCheapestPrice(4, flights, 4, 0, 3, 2)); // 300
    return 0;
}

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