Bus Routes

Hard

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You are given an array routes representing bus routes where routes[i] is a bus route that the i^th bus repeats forever.

For example, if routes[0] = [1, 5, 7], this means that the 0^th 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 <= 10⁵
All the values of routes[i] are unique.
sum(routes[i].length) <= 10⁵
0 <= routes[i][j] < 10⁶
0 <= source, target < 10⁶

Bus Routes

Hard

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Topics:

ArraysGraphs

You are given an array routes representing bus routes where routes[i] is a bus route that the i^th bus repeats forever.

For example, if routes[0] = [1, 5, 7], this means that the 0^th 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 <= 10⁵
All the values of routes[i] are unique.
sum(routes[i].length) <= 10⁵
0 <= routes[i][j] < 10⁶
0 <= source, target < 10⁶

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum number of buses and stops we can expect, and how many routes can a bus have?
Are the bus routes directed or undirected; that is, can I travel in both directions on a given bus route?
If it's impossible to reach the target stop from the starting stop, what should the function return?
Can a bus route contain duplicate stops, and if so, how should that be handled in terms of calculating the shortest path?
Are the stop IDs consecutive integers or can they be arbitrary integers, and how large could those arbitrary integers be?

Brute Force Solution

Approach

The brute force approach for finding the fewest bus transfers is like trying every possible route from your starting point to your destination. We will explore all combinations of buses to see which one requires the fewest changes. It's inefficient, but guarantees finding the absolute shortest route if one exists.

Here's how the algorithm would work step-by-step:

Start at the bus stop where you begin your journey.
Consider every single bus you can take from that starting stop.
For each of those buses, figure out all the stops that bus goes to.
From each of those stops, consider all the other buses you can transfer to.
Keep doing this, exploring all possible bus routes and transfers, until you reach your destination bus stop.
Keep track of the number of bus changes needed for each route to your destination.
Once you've explored every single possible route, pick the one that uses the fewest bus changes. That's your answer.

Code Implementation

def bus_routes_brute_force(routes, start_bus_stop, target_bus_stop):
    shortest_route = float('inf')

    def find_routes(current_stop, current_transfers, visited_routes):
        nonlocal shortest_route

        if current_stop == target_bus_stop:
            shortest_route = min(shortest_route, current_transfers)
            return

        for route_index, route in enumerate(routes):
            if current_stop in route:
                # Avoid revisiting routes to prevent infinite loops
                if route_index not in visited_routes:

                    # Explore all stops reachable from the current stop on the same route
                    for next_stop in route:
                        new_visited_routes = visited_routes | {route_index}

                        # Recursively search from this next stop with an increased transfer count
                        find_routes(next_stop, current_transfers + 1, new_visited_routes)

    # Start the search from the initial bus stop with 0 transfers
    find_routes(start_bus_stop, -1, set())

    if shortest_route == float('inf'):
        return -1
    return shortest_route

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(b^s) – In the brute force bus route approach, we are exploring every possible combination of bus routes. Let 'b' represent the average number of buses available at each stop and 's' be the number of stops we might have to visit in the worst-case scenario before reaching the target. We are branching out to 'b' different buses at each stop, and this branching can occur for up to 's' stops. Therefore, the time complexity grows exponentially with the number of stops visited and the branching factor of the buses at each stop. This leads to approximately b multiplied by itself s times operations, or b^s, which simplifies to O(b^s).

Space Complexity

O(N*M) – The described brute force approach implicitly uses a queue or stack (either explicitly or via recursion) to explore all possible bus routes. In the worst case, the algorithm might need to store information about every possible bus route. If we have N bus stops and each bus route has at most M stops, the maximum number of possible routes that need to be stored is proportional to N*M. Therefore, the auxiliary space complexity is O(N*M).

Optimal Solution

Approach

We need to find the fewest buses to get from a starting stop to a destination stop. The core idea is to treat each bus route like a connection and explore routes like how you'd explore a map. We use a technique similar to finding the shortest path to ensure we find the quickest route.

Here's how the algorithm would work step-by-step:

Start by creating a map where each bus stop is a key, and the value is a list of all the bus routes that visit that stop.
Create a way to track which routes have already been taken to prevent taking the same route more than once.
Begin at the starting bus stop and keep a record of how many bus rides it has taken to reach any given stop.
Look at all the routes that can be taken from the current stop. Mark each route as taken, and check where each route takes you.
For each stop that can be reached using the routes we've identified, determine if we've already gotten to that stop more quickly by some other route. If we have not gotten to that stop yet, or this new way of reaching the stop involves fewer bus rides, remember how many bus rides it took to get to this stop.
Keep repeating this process by selecting routes to follow and expanding outwards, level by level. Stop when you reach the destination or have explored all possible routes.
If the destination can be reached, the number of bus rides recorded is the minimum number of buses needed. If it cannot be reached, return negative one.

Code Implementation

def bus_routes(routes, source, target): 
    if source == target: 
        return 0

    stop_to_routes = {}
    for route_index, route in enumerate(routes):
        for stop in route:
            if stop not in stop_to_routes:
                stop_to_routes[stop] = []
            stop_to_routes[stop].append(route_index)

    # Keep track of visited routes to avoid cycles.
    visited_routes = set()
    queue = [(source, 0)]
    visited_stops = {source}

    while queue:
        stop, num_buses = queue.pop(0)

        # Explore all routes from the current stop.
        for route_index in stop_to_routes.get(stop, []):
            if route_index in visited_routes:
                continue

            visited_routes.add(route_index)
            
            for next_stop in routes[route_index]:
                if next_stop == target:
                    return num_buses + 1

                # Only process unvisited stops.
                if next_stop not in visited_stops:
                    visited_stops.add(next_stop)
                    queue.append((next_stop, num_buses + 1))

    return -1

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n*m) – Let n be the total number of bus stops and m be the total number of bus routes. Constructing the bus stop to routes map iterates through each stop in each route, so this is O(n*m) in the worst case where all stops are listed across all routes. The breadth-first search explores each stop and associated routes. In the worst case, we enqueue and dequeue each stop once, and for each stop, we examine all routes that visit it. The operations within the BFS contribute O(m) per stop, dominated by checking routes to adjacent stops. Thus, the BFS is O(n*m). The overall time complexity is dominated by the map construction and BFS, resulting in O(n*m).

Space Complexity

O(N) – The primary space complexity stems from the 'map where each bus stop is a key, and the value is a list of all the bus routes that visit that stop'. In the worst case, each bus stop could have its own route, and all bus routes could be unique, meaning we need to store almost all bus stops and routes. Additionally, a 'queue' would be used for the breadth-first search which, in the worst-case scenario, could contain all bus stops. Therefore, the auxiliary space is proportional to the total number of bus stops, which we'll denote as N. This gives us a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty routes array	Return -1 immediately as there are no routes to traverse.
Source and target are the same bus stop	Return 0 immediately as no bus rides are needed.
No possible route exists between source and target	Return -1 after BFS completes without finding the target.
Only one route exists, and it directly connects source and target	The BFS should find the route in the first level and return 1.
Large number of routes and stops causing potential memory issues	Use efficient data structures (e.g., adjacency lists) to minimize memory footprint.
Integer overflow with large route or stop numbers	Ensure data types used can accommodate the maximum possible stop and route values or use a different approach.
A route containing all stops	The algorithm will still process this route correctly, possibly resulting in the shortest path.
Source or target stop is not present in any of the routes	Return -1 immediately as traversal will be impossible.