Add solution #1627

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,431 LeetCode solutions in JavaScript
+# 1,432 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1255,6 +1255,7 @@
 1624|[Largest Substring Between Two Equal Characters](./solutions/1624-largest-substring-between-two-equal-characters.js)|Easy|
 1625|[Lexicographically Smallest String After Applying Operations](./solutions/1625-lexicographically-smallest-string-after-applying-operations.js)|Medium|
 1626|[Best Team With No Conflicts](./solutions/1626-best-team-with-no-conflicts.js)|Medium|
+1627|[Graph Connectivity With Threshold](./solutions/1627-graph-connectivity-with-threshold.js)|Hard|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|
 1669|[Merge In Between Linked Lists](./solutions/1669-merge-in-between-linked-lists.js)|Medium|

solutions/1627-graph-connectivity-with-threshold.js

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+/**
+ * 1627. Graph Connectivity With Threshold
+ * https://leetcode.com/problems/graph-connectivity-with-threshold/
+ * Difficulty: Hard
+ *
+ * We have n cities labeled from 1 to n. Two different cities with labels x and y are directly
+ * connected by a bidirectional road if and only if x and y share a common divisor strictly
+ * greater than some threshold. More formally, cities with labels x and y have a road between
+ * them if there exists an integer z such that all of the following are true:
+ * - x % z == 0,
+ * - y % z == 0, and
+ * - z > threshold.
+ *
+ * Given the two integers, n and threshold, and an array of queries, you must determine for each
+ * queries[i] = [ai, bi] if cities ai and bi are connected directly or indirectly. (i.e. there
+ * is some path between them).
+ *
+ * Return an array answer, where answer.length == queries.length and answer[i] is true if for
+ * the ith query, there is a path between ai and bi, or answer[i] is false if there is no path.
+ */
+
+/**
+ * @param {number} n
+ * @param {number} threshold
+ * @param {number[][]} queries
+ * @return {boolean[]}
+ */
+var areConnected = function(n, threshold, queries) {
+  const parent = new Array(n + 1).fill().map((_, i) => i);
+
+  for (let z = threshold + 1; z <= n; z++) {
+    for (let x = z; x <= n; x += z) {
+      union(x, z);
+    }
+  }
+
+  return queries.map(([a, b]) => find(a) === find(b));
+
+  function find(x) {
+    if (parent[x] !== x) {
+      parent[x] = find(parent[x]);
+    }
+    return parent[x];
+  }
+
+  function union(x, y) {
+    parent[find(x)] = find(y);
+  }
+};