<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];

        for(int i = 0; i<times.size(); i++) {
            int u = times[i][0];
            int v = times[i][1];
            int w = times[i][2];

            adj[u].push_back({v, w});
        }

        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> minheap;

        vector<int> timeTaken(n+1, 1e9);
        timeTaken[k] = 0;
        int res = 0;

        minheap.push({0, k});

        while(!minheap.empty()) {
            auto curr = minheap.top();
            int currTime = curr.first;
            int currNode = curr.second;
            minheap.pop();

            for(auto &itr: adj[currNode]) {
                int nextNode = itr[0];
                int time = itr[1];

                if(currTime + time < timeTaken[nextNode]) {
                    timeTaken[nextNode] = currTime + time;
                    minheap.push({timeTaken[nextNode], nextNode});
                    //res = max(res, timeTaken[nextNode]);
                }
            }
        }

        for(int i = 1; i<=n; i++) {
            if(timeTaken[i] == 1e9) 
                return -1;
            else
                res = max(res, timeTaken[i]);
        }
        return res;
    }
};
52290