<p>You are given a network of <code>n</code> nodes, labeled from <code>1</code> to <code>n</code>. You are also given <code>times</code>, a list of travel times as directed edges <code>times[i] = (u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>)</code>, where <code>u<sub>i</sub></code> is the source node, <code>v<sub>i</sub></code> is the target node, and <code>w<sub>i</sub></code> is the time it takes for a signal to travel from source to target.</p>

<p>We will send a signal from a given node <code>k</code>. Return <em>the <strong>minimum</strong> time it takes for all the</em> <code>n</code> <em>nodes to receive the signal</em>. If it is impossible for all the <code>n</code> nodes to receive the signal, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/23/931_example_1.png" style="width: 217px; height: 239px;" />
<pre>
<strong>Input:</strong> times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 1
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 2
<strong>Output:</strong> -1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= times.length &lt;= 6000</code></li>
	<li><code>times[i].length == 3</code></li>
	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li><code>0 &lt;= w<sub>i</sub> &lt;= 100</code></li>
	<li>All the pairs <code>(u<sub>i</sub>, v<sub>i</sub>)</code> are <strong>unique</strong>. (i.e., no multiple edges.)</li>
</ul>

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        
        vector<vector<int>> adj[n+1];
        vector<int> distances(n+1,INT_MAX);
        int cnt=0;

        for(int i=0;i<times.size();i++){
            vector<int> edge = times[i];
            int src=edge[0], dst=edge[1], time=edge[2];
            adj[src].push_back({time, dst});
        }

        // for(int i=0;i<=n;i++){
        //     cout<<i<<"m"<<adj[i].size()<<endl;
        //     for(int j=0;j<adj[i].size();j++){
        //         cout<<adj[i][j][1]<<","<<adj[i][j][0]<<" | ";
        //     }
        //     cout<<endl;
        // }

        priority_queue< vector<int>, vector< vector<int> >, greater< vector<int> > > pq;
        
        distances[k]=0;
        cnt++;
        pq.push({0,k});

        while(!pq.empty()){
            vector<int> nodeInfo = pq.top();
            pq.pop();
            int wt = nodeInfo[0], src = nodeInfo[1];

            for(int i=0;i<adj[src].size();i++){
                int time = adj[src][i][0], dst=adj[src][i][1];
                if(distances[src] + time < distances[dst]){
                    distances[dst] = distances[src] + time;
                    pq.push({distances[dst], dst});
                    cnt++;
                }
            }
        }
        int min_time = INT_MIN;
        for(int i=0;i<distances.size();i++){
            if(distances[i]>min_time && distances[i]!=INT_MAX) min_time=distances[i];
        }
        if(cnt>=n) return min_time;
        return -1;

        
    }
};
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