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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' weights along that path.</p>\n\n<p> </p>\n<p><strong>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. \nThe neighboring cities at a distanceThreshold = 4 for each city are:\nCity 0 -> [City 1, City 2] \nCity 1 -> [City 0, City 2, City 3] \nCity 2 -> [City 0, City 1, City 3] \nCity 3 -> [City 1, City 2] \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>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. \nThe neighboring cities at a distanceThreshold = 2 for each city are:\nCity 0 -> [City 1] \nCity 1 -> [City 0, City 4] \nCity 2 -> [City 3, City 4] \nCity 3 -> [City 2, City 4]\nCity 4 -> [City 1, City 2, City 3] \nThe city 0 has 1 neighboring city at a distanceThreshold = 2.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 <= n <= 100</code></li>\n\t<li><code>1 <= edges.length <= n * (n - 1) / 2</code></li>\n\t<li><code>edges[i].length == 3</code></li>\n\t<li><code>0 <= from<sub>i</sub> < to<sub>i</sub> < n</code></li>\n\t<li><code>1 <= weight<sub>i</sub>, distanceThreshold <= 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": "# @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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"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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