<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>

class Solution:
    def numBusesToDestination(self, routes: List[List[int]], source: int, target: int) -> int:
        # will be easy if we dont have route -> just stops then only need to make a graph 
        # of stops connected to each other and find the depth of BFS

        # with routes -> construct a graph 2 types of node stops and routes. stop note connect to only route and vice versa, route node connect to only stop
        graph = defaultdict(list)
        # note from 0 to n - 1 is route node
        # note from n is stops node
        if (source == target):
            return 0
            
        n = len(routes)
        for index, route in enumerate(routes):
            for stop in route:
                graph[index].append(n + stop)
                graph[n + stop].append(index)
        return self.bfs(graph, source, target, n)
    
    def bfs(self, graph, source, target, n):
        visited = set()
        queue = deque()

        queue.append(source + n)
        visited.add(source + n)

        level = len(queue)
        depth = 1

        while queue:
            # track depth
            while level > 0:
                curr = queue.popleft()
                level -= 1
                # explore neighbor:
                for neigh in graph[curr]:
                    if neigh in visited:
                        continue
                    if neigh == (target + n):
                        return depth // 2
                    queue.append(neigh)
                    visited.add(neigh)

            level = len(queue)
            depth += 1
        
        return -1

45883