<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 {
public:
    int numBusesToDestination(vector<vector<int>>& routes, int source, int target) {
                if (source == target) {
            return 0;
        }

        int num_buses = routes.size();
        
        // Step 1: Create the adjacency list for buses that share common stops
        unordered_map<int, unordered_set<int>> stop_to_buses;
        for (int bus = 0; bus < num_buses; ++bus) {
            for (int stop : routes[bus]) {
                stop_to_buses[stop].insert(bus);
            }
        }

        // Create the graph (adjacency list)
        vector<unordered_set<int>> adj(num_buses);
         for (const auto& [stop, buses] : stop_to_buses) {
            for (int bus1 : buses) {
                for (int bus2 : buses) {
                    if (bus1 != bus2) {
                        adj[bus1].insert(bus2);
                    }
                }
            }
        }

        // Step 2: BFS from all buses that have the source stop
        queue<pair<int, int>> q;  // (bus_index, buses_taken)
        vector<int> visited_buses(num_buses,0);

        for (int bus : stop_to_buses[source]) {
            q.push({bus, 1});
            visited_buses[bus]=1;
        }

        while (!q.empty()) {
            auto [current_bus, buses_taken] = q.front();
            q.pop();

            // Check if the target stop is on this bus route
            for (int stop : routes[current_bus]) {
                if (stop == target) {
                    return buses_taken;
                }
            }

            // Explore connected bus routes
            for (int next_bus : adj[current_bus]) {
                if (visited_buses[next_bus] == 0) {
                    visited_buses[next_bus]=1;
                    q.push({next_bus, buses_taken + 1});
                }
            }
        }

        return -1;  // If no route found
    }
    
};
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