Pattern this teaches:Shortest Path
Cheapest Flights Within K Stops
A medium Graph interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Graph
- Sheets
- 1
- Core for
- 12 roles
The problem
Given n cities labeled from 0 to n-1, a list of flights [from, to, price], a source city, a destination city, and a maximum number of stops k, find the cheapest price from source to destination with at most k stops. Return -1 if no such route exists.
Example 1
- Input
- n = 4, flights = [[0,1,50],[1,3,60],[0,2,80],[2,3,10]], src = 0, dst = 3, k = 1
- Output
- 90
- Why
- With one stop allowed, 0 to 2 to 3 costs 90, cheaper than 0 to 1 to 3 at 110.
Example 2
- Input
- n = 4, flights = [[0,1,50],[1,3,60],[0,2,80],[2,3,10],[0,3,150]], src = 0, dst = 3, k = 0
- Output
- 150
- Why
- With no stops allowed only the direct flight works.
Constraints
- 1 <= n <= 100
- 0 <= flights.length <= 5000
- 0 <= k < n
How to think about it
Updated 2026-09-29The constraint is not just cheapest price, but price within at most k intermediate stops (k + 1 total edges). Standard Dijkstra can discard a path with more stops that would have offered a cheaper route later. Bellman-Ford matches this edge-count bound precisely: running exactly k + 1 edge relaxation rounds restricts paths to at most k + 1 flights.
Approaches, worst first
Bellman-Ford with temporary copy array
time O(k * E) · space O(V)
Maintain an array `prices` initialized to infinity with `prices[src] = 0`. Execute k + 1 iterations: clone `prices` into `temp`, relax each flight `(u, v, p)` using `temp[v] = min(temp[v], prices[u] + p)`, and assign `prices = temp`. Return `prices[dst]` if finite, else -1.
BFS level-by-level queueWrite this one
time O(k * E) · space O(V + E)
Queue tuples of `(city, cost)`. Process level by level up to k + 1 rounds, tracking the lowest cost to each city. Prune branches if tentative cost exceeds known best cost for that city.
Where people lose marks · 3
- Updating prices in-place within the same Bellman-Ford iteration allows a chain of flights to be traversed in one round, effectively taking more than 1 flight per iteration.
- When k = 0, exactly 1 direct flight is permitted; running 0 rounds of relaxations would incorrectly leave all destinations unreachable.
- Using Dijkstra indexed purely by city without tracking current stops used leads to incorrect pruning of valid paths.
The theory behind it
Graph — the ground this problem stands on. All Graph problems
What Graph is
A graph is a network of individual points, called vertices or nodes, connected by lines called edges. Think of a subway transit map, an electrical circuit, or a web of social friends. Unlike a tree, a graph has no designated top node and no parent-child hierarchy. Connections can run one-way or both ways, and paths can loop back on themselves to form closed cycles.
When to reach for it
Reach for graph algorithms when inputs describe relationships, networks, flights between cities, course prerequisites, or clone networks. Signals include finding the shortest route across unweighted connections, ordering tasks that depend on earlier tasks, counting isolated clusters, or checking whether a path contains an infinite loop. Whenever problems present pairs of related entities and ask for reachability, distances, or dependencies, graph representations apply.
How the pattern works
First convert edge lists into an adjacency list, mapping each node to an array of its neighbors. Choose your exploration strategy based on the goal: use a queue and breadth-first search to find the shortest path in unweighted networks, or use recursion and depth-first search to explore full paths and detect cycles. Because graphs can have loops, always track visited nodes in a set or boolean array. Add nodes to the visited set at the moment they enter the queue so they are never visited twice.
More Graph problems
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