Add solution #2127

Author: JoshCrozier

README.md

@@ -449,6 +449,7 @@
 2099|[Find Subsequence of Length K With the Largest Sum](./2099-find-subsequence-of-length-k-with-the-largest-sum.js)|Medium|
 2114|[Maximum Number of Words Found in Sentences](./2114-maximum-number-of-words-found-in-sentences.js)|Easy|
 2116|[Check if a Parentheses String Can Be Valid](./2116-check-if-a-parentheses-string-can-be-valid.js)|Medium|
+2127|[Maximum Employees to Be Invited to a Meeting](./2127-maximum-employees-to-be-invited-to-a-meeting.js)|Hard|
 2129|[Capitalize the Title](./2129-capitalize-the-title.js)|Easy|
 2130|[Maximum Twin Sum of a Linked List](./2130-maximum-twin-sum-of-a-linked-list.js)|Medium|
 2154|[Keep Multiplying Found Values by Two](./2154-keep-multiplying-found-values-by-two.js)|Easy|

solutions/2127-maximum-employees-to-be-invited-to-a-meeting.js

@@ -0,0 +1,56 @@
+/**
+ * 2127. Maximum Employees to Be Invited to a Meeting
+ * https://leetcode.com/problems/maximum-employees-to-be-invited-to-a-meeting/
+ * Difficulty: Hard
+ *
+ * A company is organizing a meeting and has a list of n employees, waiting to be invited.
+ * They have arranged for a large circular table, capable of seating any number of employees.
+ *
+ * The employees are numbered from 0 to n - 1. Each employee has a favorite person and they
+ * will attend the meeting only if they can sit next to their favorite person at the table.
+ * The favorite person of an employee is not themself.
+ *
+ * Given a 0-indexed integer array favorite, where favorite[i] denotes the favorite person of
+ * the ith employee, return the maximum number of employees that can be invited to the meeting.
+ */
+
+/**
+ * @param {number[]} fav
+ * @return {number}
+ */
+var maximumInvitations = function(fav) {
+  const n = fav.length;
+  const graph = new Array(n).fill().map(() => []);
+  const seen = new Set();
+  let result = 0;
+
+  for (let i = 0; i < n; i++) {
+    graph[fav[i]].push(i);
+  }
+
+  for (let i = 0; i < n; i++) {
+    result += i < fav[i] || fav[fav[i]] != i ? 0 : dfs(i, fav[i]) + dfs(fav[i], i);
+  }
+
+  for (let i = 0; i < n; i++) {
+    let j = i;
+
+    if (!seen.has(i)) {
+      const m = new Map();
+
+      for (let k = 0; !seen.has(j); k++) {
+        seen.add(j);
+        m.set(j, k);
+        j = fav[j];
+      }
+
+      result = Math.max(result, m.size - m.get(j) || 0);
+    }
+  }
+
+  function dfs(i, j) {
+    return graph[i].reduce((max, n) => Math.max(max, n == j ? 0 : dfs(n, j)), 0) + 1;
+  };
+
+  return result;
+};