Add solution #1733

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,506 LeetCode solutions in JavaScript
+# 1,507 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1334,6 +1334,7 @@
 1727|[Largest Submatrix With Rearrangements](./solutions/1727-largest-submatrix-with-rearrangements.js)|Medium|
 1728|[Cat and Mouse II](./solutions/1728-cat-and-mouse-ii.js)|Hard|
 1732|[Find the Highest Altitude](./solutions/1732-find-the-highest-altitude.js)|Easy|
+1733|[Minimum Number of People to Teach](./solutions/1733-minimum-number-of-people-to-teach.js)|Medium|
 1748|[Sum of Unique Elements](./solutions/1748-sum-of-unique-elements.js)|Easy|
 1749|[Maximum Absolute Sum of Any Subarray](./solutions/1749-maximum-absolute-sum-of-any-subarray.js)|Medium|
 1752|[Check if Array Is Sorted and Rotated](./solutions/1752-check-if-array-is-sorted-and-rotated.js)|Easy|

solutions/1733-minimum-number-of-people-to-teach.js

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+/**
+ * 1733. Minimum Number of People to Teach
+ * https://leetcode.com/problems/minimum-number-of-people-to-teach/
+ * Difficulty: Medium
+ *
+ * On a social network consisting of m users and some friendships between users, two users can
+ * communicate with each other if they know a common language.
+ *
+ * You are given an integer n, an array languages, and an array friendships where:
+ * - There are n languages numbered 1 through n,
+ * - languages[i] is the set of languages the ith user knows, and
+ * - friendships[i] = [ui, vi] denotes a friendship between the users ui and vi.
+ *
+ * You can choose one language and teach it to some users so that all friends can communicate
+ * with each other. Return the minimum number of users you need to teach.
+ *
+ * Note that friendships are not transitive, meaning if x is a friend of y and y is a friend of
+ * z, this doesn't guarantee that x is a friend of z.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} languages
+ * @param {number[][]} friendships
+ * @return {number}
+ */
+var minimumTeachings = function(n, languages, friendships) {
+  const languageUsers = Array.from({ length: n + 1 }, () => new Set());
+  const nonCommunicating = new Set();
+
+  for (let i = 0; i < languages.length; i++) {
+    for (const lang of languages[i]) {
+      languageUsers[lang].add(i + 1);
+    }
+  }
+
+  for (const [u, v] of friendships) {
+    let canCommunicate = false;
+    for (const lang of languages[u - 1]) {
+      if (languages[v - 1].includes(lang)) {
+        canCommunicate = true;
+        break;
+      }
+    }
+    if (!canCommunicate) {
+      nonCommunicating.add(u);
+      nonCommunicating.add(v);
+    }
+  }
+
+  let result = Infinity;
+  for (let lang = 1; lang <= n; lang++) {
+    let usersToTeach = 0;
+    for (const user of nonCommunicating) {
+      if (!languages[user - 1].includes(lang)) {
+        usersToTeach++;
+      }
+    }
+    result = Math.min(result, usersToTeach);
+  }
+
+  return result;
+};