Accounts Merge

mediumcodingAlgorithmsUnion Find

Overview

Given a list of accounts where each account is [name, email1, email2, ...], merge accounts belonging to the same person. Two accounts belong to the same person if they share any email. Return merged accounts with emails sorted, name first.

Constraints

  • 1 <= accounts.length <= 1000
  • 1 <= accounts[i].length <= 10
  • All emails are valid and lowercase.

Examples

accountsMerge([
  ["John","john@mail","john_work@mail"],
  ["John","john@mail","john00@mail"],
  ["Mary","mary@mail"],
  ["John","johnny@mail"]
]);
// => [
//   ["John","john00@mail","john@mail","john_work@mail"],
//   ["John","johnny@mail"],
//   ["Mary","mary@mail"]
// ]

Solution

Reveal solution
function accountsMerge(accounts) {
  const parent = {};
  function find(x) { if (parent[x] !== x) parent[x] = find(parent[x]); return parent[x]; }
  function union(a, b) { parent[find(a)] = find(b); }

  const emailToName = {};
  for (const [name, ...emails] of accounts) {
    for (const e of emails) {
      if (!parent[e]) parent[e] = e;
      emailToName[e] = name;
      union(emails[0], e);
    }
  }

  const groups = {};
  for (const email of Object.keys(parent)) {
    const root = find(email);
    if (!groups[root]) groups[root] = [];
    groups[root].push(email);
  }

  return Object.values(groups).map(emails => {
    emails.sort();
    return [emailToName[emails[0]], ...emails];
  }).sort((a, b) => a[0].localeCompare(b[0]) || a[1].localeCompare(b[1]));
}

Resources

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