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:
- Pop the front of the queue
- If it's the destination, track the cost
- If stops exceed k, skip
- 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.
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