<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:
    struct Edge {
        int to;
        int weight;
    };
    
    int findNumReachableCities(const std::vector<std::vector<Edge>>& vertices, int distanceThreshold, int src) {
        // initialize an array with the distances to each city
        const size_t V = vertices.size();
        std::vector<int> distances(V, std::numeric_limits<int>::max());
        distances[src] = 0;

        // use a priority queue for the order that we should visit nodes
        // <distance_to_node, index>
        std::priority_queue<std::pair<int, int>> priq;
        priq.push({ distances[src], src });

        // run Djikstra's algorithm
        int num_cities = 0;
        while (!priq.empty()) {
            int distance_to_node, index;
            std::tie(distance_to_node, index) = priq.top();
            priq.pop();

            if (distances[index] < distance_to_node) {
                // stale value from the priority queue
                continue;
            }

            // update neighboring edges
            for (const Edge& e : vertices[index]) {
                int dist = distances[index] + e.weight;
                if (dist < distances[e.to] && dist <= distanceThreshold) {
                    if (distances[e.to] == std::numeric_limits<int>::max()) {
                        ++num_cities;
                    }

                    distances[e.to] = dist;
                    priq.push({ distances[e.to], e.to });
                }
            }
        }

        return num_cities;
    }

    int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
        // create adjacency list
        std::vector<std::vector<Edge>> vertices(n);
        for (const std::vector<int>& e : edges) {
            vertices[e[0]].push_back(Edge { .to = e[1], .weight = e[2] });
            vertices[e[1]].push_back(Edge { .to = e[0], .weight = e[2] });
        }

        struct Result {
            int num_cities;
            int index;
        };

        // (1) For any given city i, what is the number of cities that are reachable
        //     through some path with a distance of distanceThreshold?
        //     O(V * (V log V + E log V))
        //     std::pair<num_cities, index>
        std::vector<Result> results(n);
        for (int i = 0; i < n; ++i) {
            results[i] = { findNumReachableCities(vertices, distanceThreshold, i), i };
        }

        // (2) Find the city with the smallest number of reachable cities
        //     O(V log V)
        std::sort(results.begin(), results.end(), [](const Result& r1, const Result& r2) {
            if (r1.num_cities < r2.num_cities) {
                return true;
            }
            if (r1.num_cities > r2.num_cities) {
                return false;
            }
            return r1.index > r2.index;
        });

        return results.front().index;
    }
};
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