<p>There are <code>n</code> cities numbered from <code>0</code> to <code>n-1</code>. Given the array <code>edges</code> where <code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>, weight<sub>i</sub>]</code> represents a bidirectional and weighted edge between cities <code>from<sub>i</sub></code> and <code>to<sub>i</sub></code>, and given the integer <code>distanceThreshold</code>.</p>

<p>Return the city with the smallest number of cities that are reachable through some path and whose distance is <strong>at most</strong> <code>distanceThreshold</code>, If there are multiple such cities, return the city with the greatest number.</p>

<p>Notice that the distance of a path connecting cities <em><strong>i</strong></em> and <em><strong>j</strong></em> is equal to the sum of the edges&#39; weights along that path.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/23/problem1334example1.png" style="width: 300px; height: 224px;" /></p>

<pre>
<strong>Input:</strong> n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
<strong>Output:</strong> 3
<strong>Explanation: </strong>The figure above describes the graph.&nbsp;
The neighboring cities at a distanceThreshold = 4 for each city are:
City 0 -&gt; [City 1, City 2]&nbsp;
City 1 -&gt; [City 0, City 2, City 3]&nbsp;
City 2 -&gt; [City 0, City 1, City 3]&nbsp;
City 3 -&gt; [City 1, City 2]&nbsp;
Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/23/problem1334example0.png" style="width: 300px; height: 224px;" /></p>

<pre>
<strong>Input:</strong> n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
<strong>Output:</strong> 0
<strong>Explanation: </strong>The figure above describes the graph.&nbsp;
The neighboring cities at a distanceThreshold = 2 for each city are:
City 0 -&gt; [City 1]&nbsp;
City 1 -&gt; [City 0, City 4]&nbsp;
City 2 -&gt; [City 3, City 4]&nbsp;
City 3 -&gt; [City 2, City 4]
City 4 -&gt; [City 1, City 2, City 3]&nbsp;
The city 0 has 1 neighboring city at a distanceThreshold = 2.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= edges.length &lt;= n * (n - 1) / 2</code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= from<sub>i</sub> &lt; to<sub>i</sub> &lt; n</code></li>
	<li><code>1 &lt;= weight<sub>i</sub>,&nbsp;distanceThreshold &lt;= 10^4</code></li>
	<li>All pairs <code>(from<sub>i</sub>, to<sub>i</sub>)</code> are distinct.</li>
</ul>

// class Solution {
//     int bfs( vector<vector<pair<int,int>>>& adj,int startNode,int n, int distanceThreshold){
//      set<pair<int,int>> s;
//      s.insert({0,startNode});
//      vector<int> dist(n,INT_MAX);
//      dist[startNode] = 0;
//      vector<int> path(n,0);
//      path[startNode] = 1;
//      while(!s.empty()){
//         auto it = *s.begin();
//         s.erase(it);
//         int d = it.first;
//         int node = it.second;

//         for(auto i:adj[node]){
//             int neighbour = i.first;
//             int weight = i.second;

//             if(dist[neighbour]>d+weight){
//                 if(dist[neighbour]!=INT_MAX){
//                     s.erase({dist[neighbour],neighbour});
//                 }
//                 dist[neighbour] = d+weight;
//                 s.insert({dist[neighbour] ,neighbour});
//                 path[neighbour] = path[node];
//             }
//             else if( dist[neighbour]<=distanceThreshold){
//                 path[neighbour] += path[node];
//             }
//         }
//      }
//      return path[startNode]; 
//     }
// public:
//     int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
//        vector<vector<pair<int,int>>> adj;
//          int m = edges.size();
//         for(int i=0;i<m;i++){
//              adj[edges[i][0]].push_back({edges[i][1],edges[i][0]});
//                adj[edges[i][1]].push_back({edges[i][0],edges[i][0]});
//         }
//         return bfs(adj,0,n,distanceThreshold);
        
//     }
// };

class Solution {
public:
    int bfs(vector<vector<pair<int,int>>>& adj, int startNode, int n, int distanceThreshold) {
        set<pair<int,int>> s;
        s.insert({0, startNode});
        vector<int> dist(n, INT_MAX);
        dist[startNode] = 0;

        while (!s.empty()) {
            auto it = *s.begin();
            s.erase(it);
            int d = it.first;
            int node = it.second;
             
             if(d==distanceThreshold){
                continue;
             }
            for (auto i : adj[node]) {
                int neighbour = i.first;
                int weight = i.second;

                if (dist[neighbour] > d + weight && d+weight<=distanceThreshold) {
                    if (dist[neighbour] != INT_MAX) {
                        s.erase({dist[neighbour], neighbour});
                    }
                    dist[neighbour] = d + weight;
                    s.insert({dist[neighbour], neighbour});
                }
            
            }
        }

        int cnt =0;
        for(int i=0;i<n;i++){
            if(dist[i]!=INT_MAX && i!=startNode){
                cnt++;
            }
        }
     
        return cnt;
    }
    public:
    int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
       vector<vector<pair<int,int>>> adj(n);
         int m = edges.size();
        for(int i=0;i<m;i++){
             adj[edges[i][0]].push_back({edges[i][1],edges[i][2]});
               adj[edges[i][1]].push_back({edges[i][0],edges[i][2]});
        }

            int ans = INT_MAX;
            int minVal = INT_MAX;

        for(int i=0;i<n;i++){
            int val = bfs(adj,i,n,distanceThreshold);
            cout<<val<<" ";
            if(val<=minVal){
            minVal = val;
            ans = i;
            }
        }
        return ans;
        
    }
};

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