Find the City With the Smallest Number of Neighbors at a Threshold Distance
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1334.Find%20the%20City%20With%20the%20Smallest%20Number%20of%20Neighbors%20at%20a%20Threshold%20Distance/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1334.Find%20the%20City%20With%20the%20Smallest%20Number%20of%20Neighbors%20at%20a%20Threshold%20Distance/images/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>
Graph; Dynamic Programming; Shortest Path
class Solution {
    private int n;
    private int[][] g;
    private int[] dist;
    private boolean[] vis;
    private final int inf = 1 << 30;
    private int distanceThreshold;

    public int findTheCity(int n, int[][] edges, int distanceThreshold) {
        this.n = n;
        this.distanceThreshold = distanceThreshold;
        g = new int[n][n];
        dist = new int[n];
        vis = new boolean[n];
        for (var e : g) {
            Arrays.fill(e, inf);
        }
        for (var e : edges) {
            int f = e[0], t = e[1], w = e[2];
            g[f][t] = w;
            g[t][f] = w;
        }
        int ans = n, cnt = inf;
        for (int i = n - 1; i >= 0; --i) {
            int t = dijkstra(i);
            if (t < cnt) {
                cnt = t;
                ans = i;
            }
        }
        return ans;
    }

    private int dijkstra(int u) {
        Arrays.fill(dist, inf);
        Arrays.fill(vis, false);
        dist[u] = 0;
        for (int i = 0; i < n; ++i) {
            int k = -1;
            for (int j = 0; j < n; ++j) {
                if (!vis[j] && (k == -1 || dist[k] > dist[j])) {
                    k = j;
                }
            }
            vis[k] = true;
            for (int j = 0; j < n; ++j) {
                dist[j] = Math.min(dist[j], dist[k] + g[k][j]);
            }
        }
        int cnt = 0;
        for (int d : dist) {
            if (d <= distanceThreshold) {
                ++cnt;
            }
        }
        return cnt;
    }
}