<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>

"""
    Least number of buses you can take that take you to the destination (We don't care about the stops)

    Brute-force:
        - Find the buses that you are eligible to take from the source
        - Obtain the next stops by taking each bus
        - Find minimum buses from that stop using the remaining busses
    
    If two buses share a stop, then you can switch buses
    you need to end up at the bus which has the target stop
    you start at a bus with the source stop

    B1 <-> {B2}
    B2 <-> {B1}
    B3 <-> {}
"""

class BusRoute():
    def __init__(self, stops):
        self.stops = set(stops)
        self.links = set([])
    
    def has_stop(self, stop):
        return stop in self.stops

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

        buses = []
        stops = {}
        # O(N * M) to build this
        for bidx, r in enumerate(routes):
            for s in r:
                if s in stops:
                    stops[s].append(bidx)
                else:
                    stops[s] = [bidx]
        
        # O(N^2 x M) to link the different routes together
        for bidx, r in enumerate(routes):
            b = BusRoute(r)
            # Add all the buses you can visit along this route
            # This is O(N x M) because of set union
            other_buses = set([])
            for s in r:
                other_buses |= set(stops[s])
            b.links = other_buses
            buses.append(b)

        q = []
        # All the buses from this stop
        for bidx in stops.get(source, []):
            q.append((1, bidx))
        
        # O(N) to do a BFS since we try each bus only once
        visited = {}
        while len(q) > 0:
            dist, b = q.pop(0)
            visited[b] = True
            
            # This is O(1)
            if target in buses[b].stops:
                return dist

            for new_route in buses[b].links:
                # If you haven't taken the bus
                if visited.get(new_route, False) is False:
                    q.append((dist + 1, new_route))
        return -1
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