<p>You are given an array <code>routes</code> representing bus routes where <code>routes[i]</code> is a bus route that the <code>i<sup>th</sup></code> bus repeats forever.</p>

<ul>
	<li>For example, if <code>routes[0] = [1, 5, 7]</code>, this means that the <code>0<sup>th</sup></code> bus travels in the sequence <code>1 -&gt; 5 -&gt; 7 -&gt; 1 -&gt; 5 -&gt; 7 -&gt; 1 -&gt; ...</code> forever.</li>
</ul>

<p>You will start at the bus stop <code>source</code> (You are not on any bus initially), and you want to go to the bus stop <code>target</code>. You can travel between bus stops by buses only.</p>

<p>Return <em>the least number of buses you must take to travel from </em><code>source</code><em> to </em><code>target</code>. Return <code>-1</code> if it is not possible.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> routes = [[1,2,7],[3,6,7]], source = 1, target = 6
<strong>Output:</strong> 2
<strong>Explanation:</strong> The best strategy is take the first bus to the bus stop 7, then take the second bus to the bus stop 6.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> routes = [[7,12],[4,5,15],[6],[15,19],[9,12,13]], source = 15, target = 12
<strong>Output:</strong> -1
</pre>

<p>&nbsp;</p>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= routes.length &lt;= 500</code>.</li>
	<li><code>1 &lt;= routes[i].length &lt;= 10<sup>5</sup></code></li>
	<li>All the values of <code>routes[i]</code> are <strong>unique</strong>.</li>
	<li><code>sum(routes[i].length) &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= routes[i][j] &lt; 10<sup>6</sup></code></li>
	<li><code>0 &lt;= source, target &lt; 10<sup>6</sup></code></li>
</ul>

from collections import defaultdict, deque

class Solution:
    def numBusesToDestination(self, routes: List[List[int]], source: int, target: int) -> int:
        """
        - We can create an adjacency list from any route[i][j] to other routes.
        - We care about the shortest path in stops, not the temporally fastest path.
            So we can wait however long is necessary for a bus to come back around
        - BFS from each route?
            [1,2,7],[3,6,7]
            - Start with all routes that contain "1" -> routes[0]
            1 -> no neighbors
            2 -> no neighbors
            7 -> routes[1]
                3 -> no neighbors
                6 -> target
        - Need hashmap of stop -> routes to search
            - If route has been or is being searched, ignore it
            - Convert routes to sets so we can hash them in seen()
        - Would prefer to use references to route lists rather than make copies
            - Use frozenset
            - Double check that ids are the same
        """

        if source == target: return 0

        # stop id -> routes it's seen in
        stop_routes = defaultdict(set)
        # route id -> stops it has
        route_stops = defaultdict(set)
        for idx, route in enumerate(routes):
            for stop in route:
                stop_routes[stop].add(idx)
                route_stops[idx].add(stop)

        for stop, routes in stop_routes.items():
            stop_routes[stop] = frozenset(routes)

        for route, stops in route_stops.items():
            route_stops[route] = frozenset(stops)
 
        # routes ids to search
        queue = deque(stop_routes[source])
        next_queue = set()
        seen = set()
        dist = 1
        while queue or next_queue:
            if not queue:
                queue.extend(list(next_queue))
                next_queue = set()
                dist += 1
                if not queue: break

            search_route = queue.popleft()
            if search_route in seen: continue
            seen.add(search_route)

            # Check stops in search_route for target
            if target in route_stops[search_route]: 
                return dist

            # Search all neighbor routes
            for neighbor_stop in route_stops[search_route]:
                next_queue.update(stop_routes[neighbor_stop])

        return -1
50256