Add solution #721

Author: JoshCrozier

README.md

@@ -546,6 +546,7 @@
 718|[Maximum Length of Repeated Subarray](./0718-maximum-length-of-repeated-subarray.js)|Medium|
 719|[Find K-th Smallest Pair Distance](./0719-find-k-th-smallest-pair-distance.js)|Hard|
 720|[Longest Word in Dictionary](./0720-longest-word-in-dictionary.js)|Medium|
+721|[Accounts Merge](./0721-accounts-merge.js)|Medium|
 722|[Remove Comments](./0722-remove-comments.js)|Medium|
 724|[Find Pivot Index](./0724-find-pivot-index.js)|Easy|
 733|[Flood Fill](./0733-flood-fill.js)|Easy|

solutions/0721-accounts-merge.js

@@ -0,0 +1,49 @@
+/**
+ * 721. Accounts Merge
+ * https://leetcode.com/problems/accounts-merge/
+ * Difficulty: Medium
+ *
+ * Given a list of accounts where each element accounts[i] is a list of strings, where the first
+ * element accounts[i][0] is a name, and the rest of the elements are emails representing emails
+ * of the account.
+ *
+ * Now, we would like to merge these accounts. Two accounts definitely belong to the same person
+ * if there is some common email to both accounts. Note that even if two accounts have the same
+ * name, they may belong to different people as people could have the same name. A person can
+ * have any number of accounts initially, but all of their accounts definitely have the same name.
+ *
+ * After merging the accounts, return the accounts in the following format: the first element of
+ * each account is the name, and the rest of the elements are emails in sorted order. The accounts
+ * themselves can be returned in any order.
+ */
+
+/**
+ * @param {string[][]} accounts
+ * @return {string[][]}
+ */
+var accountsMerge = function(accounts) {
+  const parent = new Map();
+  const names = new Map();
+  const groups = new Map();
+
+  const find = node => {
+    if (!parent.has(node)) parent.set(node, node);
+    return parent.get(node) === node ? node : parent.set(node, find(parent.get(node))).get(node);
+  };
+
+  for (const [name, ...emails] of accounts) {
+    for (const email of emails) {
+      names.set(email, name);
+      find(email);
+      parent.set(find(email), find(emails[0]));
+    }
+  }
+
+  for (const email of parent.keys()) {
+    const root = find(email);
+    if (!groups.has(root)) groups.set(root, new Set());
+    groups.get(root).add(email);
+  }
+
+  return Array.from(groups, ([root, emails]) => [names.get(root), ...[...emails].sort()]);
+};