<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 {

    
vector<int> shortestPath(unordered_map<int, list<pair<int, int>>> m, int source, int n)
{
    set<pair<int, int>> st;
    vector<int> distance(n+1, INT_MAX);
    distance[source] = 0;
    st.insert(make_pair(0, source));
    while(!st.empty())
    {
        auto top = *st.begin();
        st.erase(st.begin());
        int mindistance = top.first;
        int node = top.second;
        for(auto itr: m[node])
        {
            if(itr.second + mindistance < distance[itr.first])
            {
                auto toberem = st.find(make_pair(distance[itr.first], itr.first));
                distance[itr.first] = itr.second + mindistance;
                if(toberem != st.end())
                {
                    st.erase(toberem);
                }
                st.insert(make_pair(distance[itr.first], itr.first));
            }
        }
    }
    return distance;
}

public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        unordered_map<int, list<pair<int, int>>> m;
        for(int i=0;i<times.size();i++)
        {
            m[times[i][0]].push_back({times[i][1], times[i][2]});
        }

        vector<int> distance = shortestPath(m, k, n);
        int maxi = INT_MIN;
        for(int i=1;i<distance.size();i++)
            maxi = max(maxi, distance[i]);
        if(maxi == INT_MAX)
            return -1;
        return maxi;

























    }
};
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