Add solution #1334

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,235 LeetCode solutions in JavaScript
+# 1,236 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1012,6 +1012,7 @@
 1331|[Rank Transform of an Array](./solutions/1331-rank-transform-of-an-array.js)|Easy|
 1332|[Remove Palindromic Subsequences](./solutions/1332-remove-palindromic-subsequences.js)|Easy|
 1333|[Filter Restaurants by Vegan-Friendly, Price and Distance](./solutions/1333-filter-restaurants-by-vegan-friendly-price-and-distance.js)|Medium|
+1334|[Find the City With the Smallest Number of Neighbors at a Threshold Distance](./solutions/1334-find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance.js)|Medium|
 1342|[Number of Steps to Reduce a Number to Zero](./solutions/1342-number-of-steps-to-reduce-a-number-to-zero.js)|Easy|
 1351|[Count Negative Numbers in a Sorted Matrix](./solutions/1351-count-negative-numbers-in-a-sorted-matrix.js)|Easy|
 1352|[Product of the Last K Numbers](./solutions/1352-product-of-the-last-k-numbers.js)|Medium|

solutions/1334-find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance.js

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+/**
+ * 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance
+ * https://leetcode.com/problems/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance/
+ * Difficulty: Medium
+ *
+ * There are n cities numbered from 0 to n-1. Given the array edges where
+ * edges[i] = [fromi, toi, weighti] represents a bidirectional and weighted edge between cities
+ * fromi and toi, and given the integer distanceThreshold.
+ *
+ * Return the city with the smallest number of cities that are reachable through some path and whose
+ * distance is at most distanceThreshold, If there are multiple such cities, return the city with
+ * the greatest number.
+ *
+ * Notice that the distance of a path connecting cities i and j is equal to the sum of the edges'
+ * weights along that path.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @param {number} distanceThreshold
+ * @return {number}
+ */
+var findTheCity = function(n, edges, distanceThreshold) {
+  const distances = Array.from({ length: n }, () => Array(n).fill(Infinity));
+
+  for (let i = 0; i < n; i++) {
+    distances[i][i] = 0;
+  }
+
+  for (const [from, to, weight] of edges) {
+    distances[from][to] = weight;
+    distances[to][from] = weight;
+  }
+
+  for (let k = 0; k < n; k++) {
+    for (let i = 0; i < n; i++) {
+      for (let j = 0; j < n; j++) {
+        distances[i][j] = Math.min(distances[i][j], distances[i][k] + distances[k][j]);
+      }
+    }
+  }
+
+  let minNeighbors = n;
+  let result = 0;
+
+  for (let i = 0; i < n; i++) {
+    const neighbors = distances[i].filter(dist => dist <= distanceThreshold).length - 1;
+    if (neighbors <= minNeighbors) {
+      minNeighbors = neighbors;
+      result = i;
+    }
+  }
+
+  return result;
+};