You are given an array routes representing bus routes where routes[i] is a bus route that the ith bus repeats forever.
routes[0] = [1, 5, 7], this means that the 0th bus travels in the sequence 1 -> 5 -> 7 -> 1 -> 5 -> 7 -> 1 -> ... forever.You will start at the bus stop source (You are not on any bus initially), and you want to go to the bus stop target. You can travel between bus stops by buses only.
Return the least number of buses you must take to travel from source to target. Return -1 if it is not possible.
Example 1:
Input: routes = [[1,2,7],[3,6,7]], source = 1, target = 6 Output: 2 Explanation: The best strategy is take the first bus to the bus stop 7, then take the second bus to the bus stop 6.
Example 2:
Input: routes = [[7,12],[4,5,15],[6],[15,19],[9,12,13]], source = 15, target = 12 Output: -1
Constraints:
1 <= routes.length <= 500.1 <= routes[i].length <= 105routes[i] are unique.sum(routes[i].length) <= 1050 <= routes[i][j] < 1060 <= source, target < 106Problem Overview: You are given several bus routes where each route repeats forever. Each route contains a list of bus stops. Starting from a source stop, determine the minimum number of buses you must take to reach a target stop.
Approach 1: Graph BFS on Routes (O(N * K) time, O(N * K) space)
The key idea is to treat each bus route as a node in a graph. Two routes are connected if they share at least one common stop. Instead of explicitly building the entire route graph, build a stop → list of routes mapping using a hash table. Start a Breadth-First Search from all routes containing the source stop. Each BFS level represents taking one additional bus. For every stop on the current route, look up other routes passing through that stop and push them into the queue if not visited. Stop when any route contains the target stop. This approach works well because BFS naturally finds the minimum number of bus transfers.
Approach 2: Bidirectional BFS (O(N * K) time, O(N * K) space)
Bidirectional BFS speeds up search when the graph is large. Instead of exploring only from the source, run BFS simultaneously from both the source stop and the target stop. Maintain two frontiers and expand the smaller one at each step. The same stop → routes mapping is used to discover neighboring routes. When the two searches meet at a common route or stop, you have found the minimum number of buses required. Because both sides explore fewer nodes, the practical runtime is often significantly lower than a single BFS.
Both approaches rely heavily on fast stop lookups and queue-based traversal, which makes arrays for route storage and hash maps for indexing stops the most efficient structure combination.
Recommended for interviews: The standard graph BFS approach is what most interviewers expect. It shows that you can transform the problem into a graph traversal and use BFS to compute the shortest path in terms of bus transfers. Bidirectional BFS is an optimization that demonstrates deeper graph-search knowledge and can be mentioned after presenting the core BFS solution.
In this approach, we treat the problem as a graph where each bus route is a node and bus stops are edges connecting these nodes. We will perform a Breadth-First Search (BFS) starting from the source and try to reach the target while counting the number of buses required. The BFS helps in finding the shortest path in terms of the number of bus routes.
This code uses Breadth-First Search (BFS) to explore the given routes. A queue is utilized to manage bus stops, beginning at the source, and systematically visit every possible stop until the target is achieved or all possibilities are exhausted. The unique bus stops and routes are tracked to prevent repetition. When the target stop is reached, the code returns the count of buses needed to reach that spot. If the target is unreachable, it returns -1.
Python
JavaScript
Time Complexity: O(N * D) where N is the number of buses and D is the largest number of bus stops in any bus route.
Space Complexity: O(N * D) for storing the graph.
For optimization, we can leverage a bidirectional BFS. This technique reduces the breadth by simultaneously exploring from both the source and the target and meeting in the middle. A dual directional expansion can therefore lead to faster convergence in finding the solution.
This C++ implementation uses bidirectional BFS to improve time efficiency by concurrently working from both the forward and backward directions. It alternately expands from the queue with fewer elements, using a shared graph or set to determine intersections, which indicates the optimal path. By operating in both directions, this reduces redundant processing and ensures convergence happens faster compared to one-sided BFS.
Time Complexity: O(N * D) where N is the number of buses and D is the largest number of bus stops in any bus route.
Space Complexity: O(N * D) for storing the graph and queue data structures.
First, we check if source and target are the same. If they are, we directly return 0.
Next, we use a hash table g to build a mapping from stops to bus routes. For each bus route, we traverse all the stops it passes through and map each stop to that bus route, i.e., g[stop] represents all bus routes passing through stop stop.
Then, we check if source and target are in the stop mapping. If they are not, we return -1.
We use a queue q to perform a breadth-first search (BFS). Each element in the queue is a tuple (stop, busCount), representing the current stop stop and the number of buses taken to reach the current stop busCount.
We initialize a set visBus to record the bus routes that have been visited and a set visStop to record the stops that have been visited. Then, we add source to visStop and (source, 0) to the queue q.
Next, we start the BFS. While the queue q is not empty, we take out the first element from the queue, which is the current stop stop and the number of buses taken to reach the current stop busCount.
If the current stop stop is the target stop target, we return the number of buses taken to reach the target stop busCount.
Otherwise, we traverse all bus routes passing through the current stop. For each bus route, we traverse all stops on that route. If a stop nextStop has not been visited, we add it to visStop and add (nextStop, busCount + 1) to the queue q.
Finally, if we cannot reach the target stop, we return -1.
The time complexity is O(L), and the space complexity is O(L), where L is the total number of stops on all bus routes.
| Approach | Complexity |
|---|---|
| Graph BFS Approach | Time Complexity: O(N * D) where N is the number of buses and D is the largest number of bus stops in any bus route. |
| Bidirectional BFS Approach | Time Complexity: O(N * D) where N is the number of buses and D is the largest number of bus stops in any bus route. |
| BFS | — |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Graph BFS on Routes | O(N * K) | O(N * K) | General case. Clean and widely accepted interview solution for finding minimum bus transfers. |
| Bidirectional BFS | O(N * K) | O(N * K) | Large search spaces where exploring from both source and target reduces traversal depth. |
Bus Routes || Graph theory • Pepcoding • 19,247 views views
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