<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 {
//Approach : all times are positive so we can use Dijkestra algorithm
//
private:
    struct QElement{
        int node;
        int distance;
        
        QElement(int n, int d):node(n),distance(d){}
    };
    
    struct Comprator{
        bool operator()(QElement first, QElement second){
            return first.distance>second.distance;
        }
    };
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        priority_queue<QElement, vector<QElement>, Comprator > distanceMinHeap;
        set<int> visited;
        vector<int> preStep(n+1,-1);
        vector<int> finalDistance(n+1,INT_MAX);
        int min_time=0;
        
        vector<vector<int>> dist(n+1,vector<int>(n+1,-1));
        int nTimes=times.size();
        for(int i=0;i<nTimes;++i){
            dist[times[i][0]][times[i][1]]=times[i][2];
        }
        
        //initiate priority_queue, visited, preStep
        for (int i=1;i<=n;++i){
            if (i!=k)
                distanceMinHeap.push(QElement(i,INT_MAX));            
        }
        distanceMinHeap.push(QElement(k,0));
        finalDistance[k]=0;
        preStep[k]=k;
        visited.insert(k);
        
         
        while(!distanceMinHeap.empty()){
            QElement current=distanceMinHeap.top();
            distanceMinHeap.pop();
            if(visited.find(current.node)==visited.end()){
                break;
            }
            
            finalDistance[current.node]=current.distance;
            min_time=max(min_time,current.distance);
            
            //update distanceMinHeap
            priority_queue<QElement, vector<QElement>, Comprator > temp;
            while (!distanceMinHeap.empty()){
                QElement reachingNode=distanceMinHeap.top();
                distanceMinHeap.pop();
                
                //if there is a path between current and reachin nodes:                
                if (dist[current.node][reachingNode.node]>=0){
                    visited.insert(reachingNode.node);
                    int newDist= current.distance+dist[current.node][reachingNode.node];
                    if (newDist<reachingNode.distance){
                        reachingNode.distance=newDist;
                        preStep[reachingNode.node]=current.node;
                    }
                }                
                temp.push(reachingNode);
            }
            distanceMinHeap=temp;
        }
    
        int result=visited.size()==n?min_time:-1;
        return result;         
    }
};
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