<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 Mintime(map<int,list<pair<int,int>>> adjList,int src,int n)
    {
        vector<int> dist(n+1,INT_MAX);
        set<pair<int,int>> st;
        dist[src] =0;
        st.insert({0,src});

        while(!st.empty()){
            auto topElement = *(st.begin());
            int node = topElement.second;
            int nodeDistance = topElement.first;
            st.erase(st.begin());

            for(auto nbr: adjList[node]){
                if(nodeDistance + nbr.second < dist[nbr.first]){
                    dist[nbr.first] = nodeDistance + nbr.second;
                    auto result = st.find({dist[nbr.first],nbr.first});
                    if(result!=st.end()){
                        st.erase(result);
                    }
                    st.insert({dist[nbr.first],nbr.first});
                }
            }
        }
        int ans = INT_MIN;

        for(int i=1;i<=n;i++){
            ans = max(ans,dist[i]);
        }

        return ans;
    }
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        map<int,list<pair<int,int>>> adjList;
        
        for(int i=0;i<times.size();i++){
            int u = times[i][0];
            int v = times[i][1];
            int w = times[i][2];

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

        int ans = Mintime(adjList,k,n);
        if(ans == INT_MAX){
            return -1;
        }
        return ans;
    }
};
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