721. Accounts Merge LeetCode Solution

In this guide, you will get 721. Accounts Merge LeetCode Solution with the best time and space complexity. The solution to Accounts Merge problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Accounts Merge solution in C++
4. Accounts Merge solution in Java
5. Accounts Merge solution in Python
6. Additional Resources

Problem Statement of Accounts Merge

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.

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 0) is a valid email.

Complexity Analysis

• Time Complexity: O(nk\log nk + nk\alpha(n)) = O(nk\log nk), where k = \max(|\texttt{accounts[i]}|)
• Space Complexity: O(n + nk) = O(nk)

721. Accounts Merge LeetCode Solution in C++

class UnionFind {
 public:
  UnionFind(int n) : id(n), sz(n, 1) {
    iota(id.begin(), id.end(), 0);
  }

  void unionBySize(int u, int v) {
    const int i = find(u);
    const int j = find(v);
    if (i == j)
      return;
    if (sz[i] < sz[j]) {
      sz[j] += sz[i];
      id[i] = j;
    } else {
      sz[i] += sz[j];
      id[j] = i;
    }
  }

  int find(int u) {
    return id[u] == u ? u : id[u] = find(id[u]);
  }

 private:
  vector<int> id;
  vector<int> sz;
};

class Solution {
 public:
  vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
    vector<vector<string>> ans;
    unordered_map<string, int> emailToIndex;        // {email: index}
    unordered_map<int, set<string>> indexToEmails;  // {index: {emails}}
    UnionFind uf(accounts.size());

    for (int i = 0; i < accounts.size(); ++i) {
      const string name = accounts[i][0];
      for (int j = 1; j < accounts[i].size(); ++j) {
        const string email = accounts[i][j];
        const auto it = emailToIndex.find(email);
        if (it == emailToIndex.end()) {
          emailToIndex[email] = i;
        } else {
          uf.unionBySize(i, it->second);
        }
      }
    }

    for (const auto& [email, index] : emailToIndex)
      indexToEmails[uf.find(index)].insert(email);

    for (const auto& [index, emails] : indexToEmails) {
      const string name = accounts[index][0];
      vector<string> row{name};
      row.insert(row.end(), emails.begin(), emails.end());
      ans.push_back(row);
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

721. Accounts Merge LeetCode Solution in Java

class UnionFind {
  public UnionFind(List<List<String>> accounts) {
    for (List<String> account : accounts)
      for (int i = 1; i < account.size(); ++i) {
        final String email = account.get(i);
        id.putIfAbsent(email, email);
      }
  }

  public void union(final String u, final String v) {
    id.put(find(u), find(v));
  }

  public String find(final String u) {
    if (u != id.get(u))
      id.put(u, find(id.get(u)));
    return id.get(u);
  }

  private Map<String, String> id = new HashMap<>();
}

class Solution {
  public List<List<String>> accountsMerge(List<List<String>> accounts) {
    List<List<String>> ans = new ArrayList<>();
    Map<String, String> emailToName = new HashMap<>();
    Map<String, TreeSet<String>> idEmailToEmails = new HashMap<>();
    UnionFind uf = new UnionFind(accounts);

    for (final List<String> account : accounts)
      for (int i = 1; i < account.size(); ++i)
        emailToName.putIfAbsent(account.get(i), account.get(0));

    for (final List<String> account : accounts)
      for (int i = 2; i < account.size(); ++i)
        uf.union(account.get(i), account.get(i - 1));

    for (final List<String> account : accounts)
      for (int i = 1; i < account.size(); ++i) {
        final String id = uf.find(account.get(i));
        idEmailToEmails.putIfAbsent(id, new TreeSet<>());
        idEmailToEmails.get(id).add(account.get(i));
      }

    for (final String idEmail : idEmailToEmails.keySet()) {
      List<String> emails = new ArrayList<>(idEmailToEmails.get(idEmail));
      final String name = emailToName.get(idEmail);
      emails.add(0, name);
      ans.add(emails);
    }

    return ans;
  }
}
// code provided by PROGIEZ

721. Accounts Merge LeetCode Solution in Python

N/A
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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