<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 visited[505];
    unordered_set<int> st;
    int numBusesToDestination(vector<vector<int>>& routes, int source, int target) {
        if (source == target) {
            return 0;
        }
        // # pragma unroll
        for (auto& r : routes) {
            sort(r.begin(), r.end());
        }
        
        // check if they both are on the same route
        for (int i = 0; i < routes.size(); i++) {
            int cnt = 0;
            for (int j = 0; j < routes[i].size(); j++) {
                if (routes[i][j] == source) {
                    cnt++;
                }
                if (routes[i][j] == target) {
                    cnt++;
                }
            }
            if (cnt == 2) {
                return 1;
            }
        }

        vector<int> graph[505];
        // if there a common element between two lists,
        // then we connect them with an edge
        for (int i = 0; i < routes.size(); i++) {
            for (int j = i + 1; j < routes.size(); j++) {
                // if they have a common element
                if (check(routes[i], routes[j])) {
                    graph[i].push_back(j);
                    graph[j].push_back(i);
                }
            }
        }
        // cout << "done graph\n";
        // int n = routes.size();
        // for (int i = 0; i < n; i++) {
        //     cout << "src: " << i << endl;
        //     for (int j = 0; j < graph[i].size(); j++) {
        //         cout << graph[i][j] << ", ";
        //     }
        //     cout << endl;
        // }
        int src = -1, dest = -1;
        for (int i = 0; i < routes.size(); i++) {
            for (int j = 0; j < routes[i].size(); j++) {
                if (routes[i][j] == source) {
                    src = i;
                }
                if (routes[i][j] == target) {
                    dest = i;
                }
            }
        }
        if (src == -1 or dest == -1) {
            return -1;
        }
        cout << "f " << src << " " << dest << endl;
        // run bfs from src to dest on graph
        queue<pair<int, int>> q;
        q.push({src, 0});
        int ans = 0;
        visited[src] = 1;
        while (!q.empty()) {
            int node = q.front().first;
            int d = q.front().second;
            //cout << "node aya " << node << endl;
            if (node == dest) {
                return d + 1;
            }
            q.pop();
            for (int i = 0; i < graph[node].size(); i++) {
                int neigh = graph[node][i];
                if (!visited[neigh]) {
                    visited[neigh] = 1;
                    q.push({neigh, d + 1});
                }
            }
        }
        return -1;
    }

    bool check(const vector<int>& v1, const vector<int>& v2) {
        int i = 0, j = 0;
        #pragma unroll
        while(i < v1.size() && j < v2.size()){
            if (v1[i] == v2[j]) {
                return 1;
            } else if (v1[i] < v2[j]) {
                i++;
            } else {
                j++;
            }
        }
        return 0;
    }
};
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