<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>

struct Node{
    int val;
    int distance;
    
    bool operator>(const Node& other) const {
        return distance > other.distance;
    }
};

class Solution {
public:
    int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
        
        vector<unordered_set<int>> canReach;
        unordered_map<int,vector<Node>> graph;

        for(int i=0;i<edges.size();++i) {
            graph[edges[i][0]].push_back({edges[i][1], edges[i][2]});
            graph[edges[i][1]].push_back({edges[i][0], edges[i][2]});
        }

        priority_queue<Node> pq;
        
        for(int i=0;i<n;++i) {
            canReach.push_back(dijkstra(graph, distanceThreshold, n, i));
        }

        int minSize = INT_MAX;
        int bestIdx = 0;
        
        

        for(int i=0;i<canReach.size();++i) {

            if(canReach[i].size()<=minSize) {
                minSize = canReach[i].size();
                bestIdx = max(i, bestIdx);
            }
        }

    return bestIdx;
    }

    unordered_set<int> dijkstra(unordered_map<int,vector<Node>>& graph, int distanceThreshold, int n, int start) {
        
          priority_queue<Node, vector<Node>, greater<Node>> pq;
          pq.push({start,0});
          vector<int> distance(n,INT_MAX);
          distance[start] = 0;
          unordered_set<int> res;

          while(!pq.empty()) {
            int curVal = pq.top().val;
            int curDist = pq.top().distance;
            pq.pop();

            for(auto & nei : graph[curVal]) {
                int newDist = curDist + nei.distance;
                if(distance[nei.val] > newDist) {
                    distance[nei.val] = newDist;
                    if(distance[nei.val] <= distanceThreshold) {
                        res.insert(nei.val);
                        pq.push({nei.val, distance[nei.val]});
                    }
                }
            }
          }
        //     cout << start << " distance vector" << "\n";
        //   for(int i=0;i<distance.size();++i) {
        //     cout << distance[i]<< "\n";
        //   }
        return res;
    }

};
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