<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 Dijkstra(vector<vector<int>>& edges, int n, int src){
        //first we must create an adjacency list
        vector<vector<int>> adj[n+1];
        for(int i=0; i<=edges.size()-1; i++){
            int u = edges[i][0];
            int v = edges[i][1];
            int dist = edges[i][2];
            adj[u].push_back({v, dist});
        }
        //we must use a prioirty queue and a distance vector
        vector<int> distance(n+1, INT_MAX);
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> min_heap;
        distance[src] = 0;  min_heap.push(make_pair(0, src));
        while(!min_heap.empty()){
            pair<int, int> top = min_heap.top(); min_heap.pop();
            int u = top.second;
            int dist = top.first;
            for(auto neighbour: adj[u]){
                int total_distance = dist + neighbour[1];
                if(total_distance < distance[neighbour[0]]){
                    distance[neighbour[0]] = total_distance;
                    min_heap.push(make_pair(total_distance, neighbour[0]));
                }
            }
        }
        int maxima = INT_MIN;
        for(int i=1; i<=distance.size()-1; i++){
            maxima = max(maxima, distance[i]);
        }
        if(maxima == INT_MAX) return -1;
        else return maxima;
    }
    
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(0);
        //we use Bellman Ford and Dijkstra'a single source shortest path
        //return BellmanFord(times, n, k);
        return Dijkstra(times, n, k);
    }
};
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