<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 n = routes.size();
        //we will store in diff data structure the same data
        //for easy access
        //so we will store the stops as key and route index in which which routes the current stop is present
        //so that we will deeper and find out 
        unordered_map<int,vector<int>> stops;
        for(int i=0;i<n;i++) {
            for(auto &x : routes[i]) {
                stops[x].push_back(i);
            }
        }
        vector<bool> visited(n,false);
        int levels = 0;
        queue<int> que;
        for(auto &x : stops[source]) {
            que.push(x);
        }
        while(!que.empty()) {
            int size = que.size();
            levels++;
            while(size--) {
                int curr_ind = que.front();
                que.pop();
                if(visited[curr_ind]) continue;
                vector<int> route = routes[curr_ind];
                visited[curr_ind] = true;
                for(auto &x : route) {
                    if(x == target) return levels;
                    for(auto &r : stops[x]) {
                        if(!visited[r]) {
                            que.push(r);
                        }
                    }
                }
            }
        }
        return -1;
    }
};
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