<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>

#define PII pair<int, pair<int,int>>
#define PI pair<int,int>
class Comparator{
    public:
    bool operator()(const PI &a, const PI &b){
        return a.first < b.first;

      }
};
class Solution {
public:
    map<int, set<int>> ans;
    vector<PI>G[105];
    void dij(int node, int thresh, int n) {
        int vis[102] = {0};
        int dis[102] = {0};
        vis[node] = 1;
        for(int i =0; i<= n; i++) dis[i] = INT_MAX;
        dis[node] = 0;
        priority_queue<PI, vector<PI>, Comparator> q;
        q.push(make_pair(0, node));
        while(!q.empty()) {
            PI p = q.top();
            q.pop();
            int w = p.first;
            int u = p.second;
            int sz = G[u].size();
            for (int i = 0; i < sz; i++) {
                int u1, v1, w1;
                w1 = G[u][i].second + w;
                v1 = G[u][i].first;
                if (  w1 <= thresh && w1 < dis[v1]) {
                    dis[v1] = w1;
                    q.push(make_pair(w1, v1));
                    this->ans[node].insert(v1);
                    vis[v1] = 1;
                }

            }
        }
        // cout << node << " node : " << endl;
        // for(int i =0; i < n; i++) cout << dis[i] << " ";
        // cout << endl;
        // int sz = this->G[node].size();
    }

    int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
        int ln = edges.size();
        for(int i = 0; i < ln; i++) {
            int u = edges[i][0];
            int v = edges[i][1];
            int w = edges[i][2];
            G[u].push_back(make_pair(v,w));
            G[v].push_back(make_pair(u,w));
        }
        for(int i = 0 ; i < n; i++) {
            set<int> jj;
            if(this->ans.find(i) == this->ans.end()) {
                this->ans[i] = jj;
            }
            dij(i, distanceThreshold, n);
        }

        map<int, set<int>> ::iterator it;
        int node_ans = INT_MAX, idx = -1;
        for(it = this->ans.begin(); it != ans.end(); ++it) {
            // cout << it->first << " " << it->second.size() << endl;
            if(it->second.size() == node_ans) {
                node_ans = it->second.size();
                idx = max(it->first, idx);
            }
            if(it->second.size() < node_ans) {
                node_ans = it->second.size();
                idx = it->first;
            } 
        }

        
    return idx;
        
    }
};
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