<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 {
              // city      to city | weight
    unordered_map<int,unordered_map<int,int>> adjList;
    unordered_map<int,set<int>> reach;
public:
    int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
        int answer = 0;
        buildAdjList ( edges);
        for ( int i = 0; i < n; ++i) {
            int dist = 0;
            unordered_map<int,int> visited;
            dfs ( i, distanceThreshold, dist, i, visited);
        }
        int minCities = INT_MAX;
        for ( int i = 0; i < n; ++i) {
            minCities = min ( (size_t)minCities, reach[i].size());
            if ( minCities == reach[i].size() )
                answer = i;
        }
        return answer;
    }
    void buildAdjList ( vector<vector<int>>& edges) {
        for ( vector<int> & conn : edges) {
            int from = conn[0];
            int to   = conn[1];
            int w   = conn[2];
            adjList[from][to] = w;
            adjList[to][from] = w;
        }
    }
    void dfs ( int city, int distanceThreshold, int dist, int rootCity, unordered_map<int,int> & visited) {
        unordered_map<int,int> & citiesAround = adjList[city];
        visited[city] = dist;
        for ( auto & entry : citiesAround) {
            int currentW = dist + entry.second;
            if ( visited.count(entry.first) == 0 || visited[entry.first] > currentW) {                            
                if ( currentW <= distanceThreshold)
                {
                    reach[rootCity].insert(entry.first);
                    dfs( entry.first, distanceThreshold, dist + entry.second, rootCity, visited  );
                }
            }             
        }
    }
};
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