721. Accounts Merge

Problem

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.

https://leetcode.com/problems/accounts-merge/

Example 1:

Input: accounts = [["John","johnsmith@mail.com","john_newyork@mail.com"],["John","johnsmith@mail.com","john00@mail.com"],["Mary","mary@mail.com"],["John","johnnybravo@mail.com"]]
Output: [["John","john00@mail.com","john_newyork@mail.com","johnsmith@mail.com"],["Mary","mary@mail.com"],["John","johnnybravo@mail.com"]]
Explanation:
The first and second John’s are the same person as they have the common email "johnsmith@mail.com".
The third John and Mary are different people as none of their email addresses are used by other accounts.
We could return these lists in any order, for example the answer [['Mary', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], ['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.

Example 2:

Input: accounts = [["Gabe","Gabe0@m.co","Gabe3@m.co","Gabe1@m.co"],["Kevin","Kevin3@m.co","Kevin5@m.co","Kevin0@m.co"],["Ethan","Ethan5@m.co","Ethan4@m.co","Ethan0@m.co"],["Hanzo","Hanzo3@m.co","Hanzo1@m.co","Hanzo0@m.co"],["Fern","Fern5@m.co","Fern1@m.co","Fern0@m.co"]]
Output: [["Ethan","Ethan0@m.co","Ethan4@m.co","Ethan5@m.co"],["Gabe","Gabe0@m.co","Gabe1@m.co","Gabe3@m.co"],["Hanzo","Hanzo0@m.co","Hanzo1@m.co","Hanzo3@m.co"],["Kevin","Kevin0@m.co","Kevin3@m.co","Kevin5@m.co"],["Fern","Fern0@m.co","Fern1@m.co","Fern5@m.co"]]

Constraints:

1 <= accounts.length <= 1000
2 <= accounts[i].length <= 10
1 <= accounts[i][j].length <= 30
accounts[i][0] consists of English letters.
accounts[i][j] (for j > 0) is a valid email.

Test Cases

1 2	class Solution: def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:

solution_test.py

import pytest

from solution import Solution


@pytest.mark.parametrize('accounts, expected', [
    (
        [["John","johnsmith@mail.com","john_newyork@mail.com"],["John","johnsmith@mail.com","john00@mail.com"],["Mary","mary@mail.com"],["John","johnnybravo@mail.com"]],
        [["John","john00@mail.com","john_newyork@mail.com","johnsmith@mail.com"],["Mary","mary@mail.com"],["John","johnnybravo@mail.com"]]
    ),
    (
        [["Gabe","Gabe0@m.co","Gabe3@m.co","Gabe1@m.co"],["Kevin","Kevin3@m.co","Kevin5@m.co","Kevin0@m.co"],["Ethan","Ethan5@m.co","Ethan4@m.co","Ethan0@m.co"],["Hanzo","Hanzo3@m.co","Hanzo1@m.co","Hanzo0@m.co"],["Fern","Fern5@m.co","Fern1@m.co","Fern0@m.co"]],
        [["Ethan","Ethan0@m.co","Ethan4@m.co","Ethan5@m.co"],["Gabe","Gabe0@m.co","Gabe1@m.co","Gabe3@m.co"],["Hanzo","Hanzo0@m.co","Hanzo1@m.co","Hanzo3@m.co"],["Kevin","Kevin0@m.co","Kevin3@m.co","Kevin5@m.co"],["Fern","Fern0@m.co","Fern1@m.co","Fern5@m.co"]]
    ),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, accounts, expected):
    assert sorted(sol.accountsMerge(accounts)) == sorted(expected)

Thoughts

以 accounts 中的每条记录为顶点作一张图，如果两个 accounts 的 emails 中有相同的，则把这两个顶点用边连起来。最终连在一起的 account 就都是同一个人的。

684. Redundant Connection 中提到的 disjoint set 很适合做这个。先维护一个 email 到 account id（accounts 数组的下标）的字典，如果一个 email 同时属于多个 accounts，则利用 disjoint set 合并这两个 accounts。

最后利用 disjoint set，找到每个 email 所归属的连在一起的 accounts 中的一个（树根）就行了。

Code

solution.py

from collections import defaultdict


class DisjointSet:
    def __init__(self, n: int) -> None:
        self.parent = list(range(n)) # parent[u]: u's parent node.
        self.depth = [0] * n # depth[u]: the max depth starting from u.

    def find(self, u: int) -> int:
        while self.parent[u] != u: u = self.parent[u]
        return u

    def union(self, u: int, v: int) -> None:
        ur = self.find(u)
        vr = self.find(v)
        # if ur == vr: return False

        if (diff := self.depth[ur] - self.depth[vr]) == 0:
            self.depth[ur] += 1
        elif diff < 0:
            ur, vr = vr, ur # Make sure that depth[ur] >= depth[vr]

        self.parent[vr] = ur
        # return True


class Solution:
    def accountsMerge(self, accounts: list[list[str]]) -> list[list[str]]:
        indexing: dict[str, int] = {} # {email: account id}
        disjoint = DisjointSet(len(accounts))
        for uid, emails in enumerate(accounts):
            for j in range(1, len(emails)):
                email = emails[j]
                if email not in indexing:
                    indexing[email] = uid
                else:
                    disjoint.union(uid, indexing[email])

        merged: dict[int, list[str]] = defaultdict(list)
        for email, uid in indexing.items():
            uid = disjoint.find(uid)
            merged[uid].append(email)

        return [[accounts[uid][0], *sorted(emails)] for uid, emails in merged.items()]