Cheapest Flights Within K Stops

Description:​

There are n cities connected by some number of flights. You are given an array flights where flights[i] = [fromi, toi, pricei] indicates that there is a flight from city fromi to city toi with cost pricei.

You are also given three integers src, dst, and k, return the cheapest price from src to dst with at most k stops. If there is no such route, return -1.

Video Solution:​

Approaches:​

1. Breadth-First Search (Modified Dijkstra) (Optimal)​

Unlike standard shortest-path algorithms like Dijkstra's (which prioritizes by minimum cost using a Priority Queue), this problem imposes a strict restriction on the number of stops (k). If we prioritize purely by cost, we might reach a city cheaply but exhaust our limit of stops, missing a slightly more expensive route that uses fewer stops and can successfully reach the destination.

Since the number of stops increments by exactly 1 with each flight taken, Breadth-First Search (BFS) is the ideal approach. By using a standard Queue where elements are ordered naturally by the number of stops, we explore paths level-by-level (0 stops, 1 stop, ..., up to k stops).

Algorithm:

1. Build an adjacency list adj from the flights array where adj[u] contains pairs (v, price).
2. Initialize a minCost array of size n with infinity (or a very large number), setting minCost[src] = 0.
3. Use a queue that stores tuples of (stops, current_city, current_cost). Push the initial state (0, src, 0) into the queue.
4. While the queue is not empty:
   • Pop the front element (stops, u, cost).
   • If stops > k, we cannot take any more flights from this city, so continue to the next item in the queue.
   • For each neighbor (v, price) in adj[u]:
     • Calculate newCost = cost + price.
     • If newCost < minCost[v], update minCost[v] = newCost and push (stops + 1, v, newCost) into the queue.
5. After the queue is empty, if minCost[dst] is still infinity, return -1. Otherwise, return minCost[dst].

Complexity​

• Time Complexity: $O(V + E)$ where $V$ is the number of cities (n) and $E$ is the number of flights (flights.length). Since we explore level-by-level up to at most k stops and only push neighbors into the queue when a cheaper path is found, each edge is processed efficiently without redundant cycles.
• Space Complexity: $O(V + E)$ to store the adjacency list graph, the minCost array, and the queue elements.

Solutions:​

C++

class Solution {
public:
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int k) {
        // Build adjacency list: adj[u] = {v, price}
        vector<vector<pair<int, int>>> adj(n);
        for (const auto& flight : flights) {
            adj[flight[0]].push_back({flight[1], flight[2]});
        }
        
        // minCost array initialized to infinity
        vector<int> minCost(n, 1e9);
        minCost[src] = 0;
        
        // Queue stores: {stops, {current_node, current_cost}}
        queue<pair<int, pair<int, int>>> q;
        q.push({0, {src, 0}});
        
        while (!q.empty()) {
            auto [stops, nodeCost] = q.front();
            auto [u, cost] = nodeCost;
            q.pop();
            
            // If stops exceed limit k, stop exploring further from this path
            if (stops > k) continue;
            
            for (const auto& [v, price] : adj[u]) {
                if (cost + price < minCost[v]) {
                    minCost[v] = cost + price;
                    q.push({stops + 1, {v, minCost[v]}});
                }
            }
        }
        
        return minCost[dst] == 1e9 ? -1 : minCost[dst];
    }
};

Java

class Solution {
    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int k) {
        // Build adjacency list: adj[u] = list of int[]{v, price}
        List<List<int[]>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int[] flight : flights) {
            adj.get(flight[0]).add(new int[]{flight[1], flight[2]});
        }
        
        // minCost array initialized to infinity
        int[] minCost = new int[n];
        Arrays.fill(minCost, Integer.MAX_VALUE);
        minCost[src] = 0;
        
        // Queue stores: int[]{stops, current_node, current_cost}
        Queue<int[]> q = new LinkedList<>();
        q.offer(new int[]{0, src, 0});
        
        while (!q.isEmpty()) {
            int[] current = q.poll();
            int stops = current[0];
            int u = current[1];
            int cost = current[2];
            
            if (stops > k) continue;
            
            for (int[] neighbor : adj.get(u)) {
                int v = neighbor[0];
                int price = neighbor[1];
                
                if (cost + price < minCost[v]) {
                    minCost[v] = cost + price;
                    q.offer(new int[]{stops + 1, v, minCost[v]});
                }
            }
        }
        
        return minCost[dst] == Integer.MAX_VALUE ? -1 : minCost[dst];
    }
}

Python

from collections import deque

class Solution:
    def findCheapestPrice(self, n: int, flights: list[list[int]], src: int, dst: int, k: int) -> int:
        # Build adjacency list: adj[u] = [(v, price)]
        adj = [[] for _ in range(n)]
        for u, v, price in flights:
            adj[u].append((v, price))
            
        # minCost array initialized to infinity
        minCost = [float('inf')] * n
        minCost[src] = 0
        
        # Queue stores tuples of: (stops, current_node, current_cost)
        q = deque([(0, src, 0)])
        
        while q:
            stops, u, cost = q.popleft()
            
            if stops > k:
                continue
                
            for v, price in adj[u]:
                if cost + price < minCost[v]:
                    minCost[v] = cost + price
                    q.append((stops + 1, v, minCost[v]))
                    
        return minCost[dst] if minCost[dst] != float('inf') else -1

JavaScript

/**
 * @param {number} n
 * @param {number[][]} flights
 * @param {number} src
 * @param {number} dst
 * @param {number} k
 * @return {number}
 */
var findCheapestPrice = function(n, flights, src, dst, k) {
    // Build adjacency list: adj[u] = [[v, price]]
    const adj = Array.from({ length: n }, () => []);
    for (const [u, v, price] of flights) {
        adj[u].push([v, price]);
    }
    
    // minCost array initialized to infinity
    const minCost = new Array(n).fill(Infinity);
    minCost[src] = 0;
    
    // Queue stores: [stops, current_node, current_cost]
    const q = [[0, src, 0]];
    let head = 0; // Using pointer to avoid array shift O(N) overhead
    
    while (head < q.length) {
        const [stops, u, cost] = q[head++];
        
        if (stops > k) continue;
        
        for (const [v, price] of adj[u]) {
            if (cost + price < minCost[v]) {
                minCost[v] = cost + price;
                q.push([stops + 1, v, minCost[v]]);
            }
        }
    }
    
    return minCost[dst] === Infinity ? -1 : minCost[dst];
};