815. Bus Routes - Explanation

Description

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

For example, if 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 <= 1,00,000
All the values of routes[i] are unique.
sum(routes[i].length) <= 1,00,000
0 <= routes[i][j] < 1,000,000
0 <= source, target < 1,000,000

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1. Breadth First Search (Stops as Nodes)

class Solution:
    def numBusesToDestination(self, routes: List[List[int]], source: int, target: int) -> int:
        if source == target:
            return 0
        
        n = len(routes)
        stops = defaultdict(list)
        for bus in range(n):
            for stop in routes[bus]:
                stops[stop].append(bus)
        
        seen_bus = set()
        seen_stop = set([source])
        res = 0
        q = deque([source])
        while q:
            for _ in range(len(q)):
                stop = q.popleft()
                if stop == target:
                    return res
                for bus in stops[stop]:
                    if bus in seen_bus:
                        continue
                    seen_bus.add(bus)
                    for nxtStop in routes[bus]:
                        if nxtStop in seen_stop:
                            continue
                        seen_stop.add(nxtStop)
                        q.append(nxtStop)
            res += 1
        
        return -1

Time & Space Complexity

Time complexity: $O(n * m)$
Space complexity: $O(n * m)$

Where $n$ is the number of routes and $m$ is the maximum number of stops per bus route.

2. Breadth First Search (Routes as Nodes)

class Solution:
    def numBusesToDestination(self, routes: List[List[int]], source: int, target: int) -> int:
        if source == target:
            return 0

        n = len(routes)
        adjList = [[] for _ in range(n)]
        stopToRoutes = defaultdict(list)
        for bus, route in enumerate(routes):
            for stop in route:
                stopToRoutes[stop].append(bus)

        if target not in stopToRoutes or source not in stopToRoutes:
            return -1

        hasEdge = [[False] * n for _ in range(n)]
        for buses in stopToRoutes.values():
            for i in range(len(buses)):
                for j in range(i + 1, len(buses)):
                    if hasEdge[buses[i]][buses[j]]:
                        continue
                    hasEdge[buses[i]][buses[j]] = True
                    hasEdge[buses[j]][buses[i]] = True
                    adjList[buses[i]].append(buses[j])
                    adjList[buses[j]].append(buses[i])

        q = deque([node for node in stopToRoutes[source]])
        res = 1
        while q:
            for _ in range(len(q)):
                node = q.popleft()
                if node in stopToRoutes[target]:
                    return res
                while adjList[node]:
                    nxtBus = adjList[node].pop()
                    if adjList[nxtBus]:
                        q.append(nxtBus)
            res += 1

        return -1

Time & Space Complexity

Time complexity: $O(n ^ 2 + n * m)$
Space complexity: $O(n ^ 2 + n * m)$

Where $n$ is the number of routes and $m$ is the maximum number of stops per bus route.