<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>

struct comparator{
    constexpr bool operator()(vector<int> const & a, vector<int> const & b)noexcept{
        return a[1] > b[1];
    }
};
using my_priority_queue = priority_queue<vector<int>, vector<vector<int>>, comparator>;
class Solution {
public:
    void buildAdj(vector<vector<int>>& adj, vector<vector<int>> times, int n){
        for(vector<int> time : times){
            adj[time[0]-1][time[1]-1] = time[2];
        }
    }
    int dijkstra(vector<vector<int>>& adj, vector<int>& dist, vector<bool>& visited, int k){
        //pq will hold a b pair a = node b = distance
        my_priority_queue pq;
        //initialize with startPoint
        dist[k-1] = 0;
        pq.push({k-1,0});

        while(!pq.empty()){
            int node = pq.top()[0];
            int currDist = pq.top()[1];
            //cout << node << " "  << currDist << endl;
            pq.pop();
            if(!visited[node]){
            visited[node] = true;
            //update neighbors
            for(int neighbor = 0; neighbor < adj.size(); neighbor++){
                if(adj[node][neighbor]  != -1 && !visited[neighbor]){
                    int minDist = min(dist[neighbor], currDist + adj[node][neighbor]);
                    dist[neighbor] = minDist;
                    pq.push({neighbor, minDist});
                }
            }
            }
        }
        return (find(dist.begin(), dist.end(), INT_MAX) != dist.end()) ? -1 : *max_element(dist.begin(), dist.end());
    }
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        vector<bool> visited(n, false);
        vector<int> dist(n, INT_MAX);

        // build Adj matrix?
        vector<vector<int>> adj(n, vector<int>(n,-1));
        buildAdj(adj, times, n);
        
        return dijkstra(adj, dist, visited, k);
    }
};
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