<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 {
public:
    int getMinReached(int i,int n, vector<pair<int,int>> *adj, int distanceThreshold) {
        if(distanceThreshold == 0) return 0;
        vector<int> visited(n,INT_MAX);
        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> q; // { dist covered so far, node}
        q.push({0, i});
        int cityReached = 0; // curr city
        while(!q.empty()) {
            auto node = q.top(); q.pop();
            int curr = node.second;
            int dist = node.first;
            for(auto nextNeigh: adj[curr]) {
                if(dist + nextNeigh.second <= distanceThreshold && visited[nextNeigh.first] > dist + nextNeigh.second) {
                    q.push({dist + nextNeigh.second, nextNeigh.first});
                }
            }
            if(visited[curr] == INT_MAX) {
                cout<<" curr "<<curr<<" dist for far "<<dist<<endl;
                visited[curr] = dist;
                cityReached ++;
            }
        }

        return cityReached;
    }
    int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
        vector<pair<int,int>> adj[n];
        for(int i=0;i<edges.size();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 city = -1;
        int minReached = INT_MAX;

        for(int i=0;i<n;i++) {
            int reached = getMinReached(i, n, adj, distanceThreshold);
            cout<<"reached "<<reached<<" node "<<i<<endl;
            if(reached <= minReached) {
                minReached = reached;
                city = i;
            }
        }

        return city;
    }
};
/*
0-1 2
0-4 8
1-2 3
1-4 2
2-3 1
3-4 1
dist thres = 2
0 - {1,2}, {4,8}=> {1}
1 - {0,2}, {2,3}, {4,2} => {0,4}
2 - {1,3}, {3,1} => {3, 4}
3 - {2,1}, {4,1} => {2}
4 - {0,8}, {1,2}, {3,1}

*/
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