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leetcode-problemset/leetcode/originData/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance.json

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{
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"title": "Find the City With the Smallest Number of Neighbors at a Threshold Distance",
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"content": "<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>\n\n<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>\n\n<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>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/01/16/find_the_city_01.png\" style=\"width: 300px; height: 225px;\" />\n<pre>\n<strong>Input:</strong> n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4\n<strong>Output:</strong> 3\n<strong>Explanation: </strong>The figure above describes the graph.&nbsp;\nThe neighboring cities at a distanceThreshold = 4 for each city are:\nCity 0 -&gt; [City 1, City 2]&nbsp;\nCity 1 -&gt; [City 0, City 2, City 3]&nbsp;\nCity 2 -&gt; [City 0, City 1, City 3]&nbsp;\nCity 3 -&gt; [City 1, City 2]&nbsp;\nCities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/01/16/find_the_city_02.png\" style=\"width: 300px; height: 225px;\" />\n<pre>\n<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\n<strong>Output:</strong> 0\n<strong>Explanation: </strong>The figure above describes the graph.&nbsp;\nThe neighboring cities at a distanceThreshold = 2 for each city are:\nCity 0 -&gt; [City 1]&nbsp;\nCity 1 -&gt; [City 0, City 4]&nbsp;\nCity 2 -&gt; [City 3, City 4]&nbsp;\nCity 3 -&gt; [City 2, City 4]\nCity 4 -&gt; [City 1, City 2, City 3]&nbsp;\nThe city 0 has 1 neighboring city at a distanceThreshold = 2.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 100</code></li>\n\t<li><code>1 &lt;= edges.length &lt;= n * (n - 1) / 2</code></li>\n\t<li><code>edges[i].length == 3</code></li>\n\t<li><code>0 &lt;= from<sub>i</sub> &lt; to<sub>i</sub> &lt; n</code></li>\n\t<li><code>1 &lt;= weight<sub>i</sub>,&nbsp;distanceThreshold &lt;= 10^4</code></li>\n\t<li>All pairs <code>(from<sub>i</sub>, to<sub>i</sub>)</code> are distinct.</li>\n</ul>\n",
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"code": "class Solution {\npublic:\n int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {\n \n }\n};",
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"code": "int findTheCity(int n, int** edges, int edgesSize, int* edgesColSize, int distanceThreshold) {\n \n}",
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"code": "/**\n * @param {number} n\n * @param {number[][]} edges\n * @param {number} distanceThreshold\n * @return {number}\n */\nvar findTheCity = function(n, edges, distanceThreshold) {\n \n};",
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"code": "class Solution {\n\n /**\n * @param Integer $n\n * @param Integer[][] $edges\n * @param Integer $distanceThreshold\n * @return Integer\n */\n function findTheCity($n, $edges, $distanceThreshold) {\n \n }\n}",
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"lang": "Kotlin",
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"lang": "Ruby",
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"code": "# @param {Integer} n\n# @param {Integer[][]} edges\n# @param {Integer} distance_threshold\n# @return {Integer}\ndef find_the_city(n, edges, distance_threshold)\n \nend",
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"code": "object Solution {\n def findTheCity(n: Int, edges: Array[Array[Int]], distanceThreshold: Int): Int = {\n \n }\n}",
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"lang": "Rust",
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"code": "impl Solution {\n pub fn find_the_city(n: i32, edges: Vec<Vec<i32>>, distance_threshold: i32) -> i32 {\n \n }\n}",
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"code": "(define/contract (find-the-city n edges distanceThreshold)\n (-> exact-integer? (listof (listof exact-integer?)) exact-integer? exact-integer?)\n )",
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"code": "-spec find_the_city(N :: integer(), Edges :: [[integer()]], DistanceThreshold :: integer()) -> integer().\nfind_the_city(N, Edges, DistanceThreshold) ->\n .",
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"code": "defmodule Solution do\n @spec find_the_city(n :: integer, edges :: [[integer]], distance_threshold :: integer) :: integer\n def find_the_city(n, edges, distance_threshold) do\n \n end\nend",
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"hints": [
"Use Floyd-Warshall's algorithm to compute any-point to any-point distances. (Or can also do Dijkstra from every node due to the weights are non-negative).",
"For each city calculate the number of reachable cities within the threshold, then search for the optimal city."
],
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