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‎README.md

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| # | Title | Solutions | Time | Space | Difficulty | Tag | Notes
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|-----|----------------|---------------|---------------|---------------|-------------|--------------|-----
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|721|[Accounts Merge](https://leetcode.com/problems/accounts-merge/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_721.java) | | | Medium | DFS, Union Find
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|720|[Longest Word in Dictionary](https://leetcode.com/problems/longest-word-in-dictionary/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_720.java) | O(∑wi) where wi is the length of words[i] | O(∑wi) where wi is the length of words[i] | Easy | Trie
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|719|[Find K-th Smallest Pair Distance](https://leetcode.com/problems/find-k-th-smallest-pair-distance/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_719.java) | O(nlogw + nlogn) | O(1) | Hard | Binary Search
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|718|[Maximum Length of Repeated Subarray](https://leetcode.com/problems/maximum-length-of-repeated-subarray/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_718.java) | O(m*n) | O(m*n) | Medium | DP

‎src/main/java/com/fishercoder/solutions/_721.java

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package com.fishercoder.solutions;
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import java.util.ArrayList;
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import java.util.Collections;
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import java.util.HashMap;
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import java.util.HashSet;
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import java.util.List;
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import java.util.Map;
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import java.util.Set;
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import java.util.Stack;
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/**
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* 721. Accounts Merge
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*
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* Given a list accounts, 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.
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* Now, we would like to merge these accounts.
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* Two accounts definitely belong to the same person if there is some email that is common to both accounts.
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* Note that even if two accounts have the same name, they may belong to different people as people could have the same name.
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* A person can have any number of accounts initially, but all of their accounts definitely have the same name.
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* After merging the accounts, return the accounts in the following format:
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* the first element of each account is the name, and the rest of the elements are emails in sorted order.
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* The accounts themselves can be returned in any order.
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Example 1:
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Input:
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accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
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Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'], ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
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Explanation:
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The first and third John's are the same person as they have the common email "johnsmith@mail.com".
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The second John and Mary are different people as none of their email addresses are used by other accounts.
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We could return these lists in any order, for example the answer [['Mary', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'],
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['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.
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Note:
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The length of accounts will be in the range [1, 1000].
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The length of accounts[i] will be in the range [1, 10].
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The length of accounts[i][j] will be in the range [1, 30].
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*/
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public class _721 {
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public static class Solution1 {
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/**credit: https://leetcode.com/articles/accounts-merge/#approach-1-depth-first-search-accepted
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*
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* Time Complexity: O(∑ai*logai) where a​i is the length of accounts[i].
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* Without the log factor, this is the complexity to build the graph and search for each component. The log factor is for sorting each component at the end.
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* Space Complexity: O(∑ai) the space used by the graph and search.
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* .*/
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public List<List<String>> accountsMerge(List<List<String>> accounts) {
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Map<String, String> emailToName = new HashMap();
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Map<String, ArrayList<String>> graph = new HashMap();
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for (List<String> account : accounts) {
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String name = "";
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for (String email : account) {
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if (name == "") {
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name = email;
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continue;
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}
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graph.computeIfAbsent(email, x -> new ArrayList<>()).add(account.get(1));
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graph.computeIfAbsent(account.get(1), x -> new ArrayList<>()).add(email);
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emailToName.put(email, name);
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}
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}
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Set<String> seen = new HashSet();
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List<List<String>> ans = new ArrayList();
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for (String email : graph.keySet()) {
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if (!seen.contains(email)) {
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seen.add(email);
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Stack<String> stack = new Stack();
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stack.push(email);
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List<String> component = new ArrayList();
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while (!stack.empty()) {
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String node = stack.pop();
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component.add(node);
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for (String nei : graph.get(node)) {
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if (!seen.contains(nei)) {
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seen.add(nei);
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stack.push(nei);
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}
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}
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}
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Collections.sort(component);
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component.add(0, emailToName.get(email));
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ans.add(component);
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}
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}
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return ans;
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}
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}
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public static class Solution2 {
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/**credit: https://leetcode.com/articles/accounts-merge/#approach-2-union-find-accepted
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* DSU stands for Disjoint Set Union: https://en.wikipedia.org/wiki/Disjoint-set_data_structure
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*
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* Time complexity: O(AlogA)
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* Space complexity: O(A)
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* */
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public List<List<String>> accountsMerge(List<List<String>> accounts) {
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DSU dsu = new DSU();
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Map<String, String> emailToName = new HashMap<>();
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Map<String, Integer> emailToId = new HashMap<>();
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int id = 0;
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for (List<String> account : accounts) {
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String name = "";
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for (String email : account) {
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if (name.equals("")) {
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name = email;
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continue;
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}
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emailToName.put(email, name);
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if (!emailToId.containsKey(email)) {
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emailToId.put(email, id++);
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}
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dsu.union(emailToId.get(account.get(1)), emailToId.get(email));
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}
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}
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Map<Integer, List<String>> ans = new HashMap<>();
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for (String email : emailToName.keySet()) {
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int index = dsu.find(emailToId.get(email));
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ans.computeIfAbsent(index, x -> new ArrayList()).add(email);
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}
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for (List<String> component : ans.values()) {
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Collections.sort(component);
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component.add(0, emailToName.get(component.get(0)));
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}
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return new ArrayList<>(ans.values());
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}
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class DSU {
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int[] parent;
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public DSU() {
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parent = new int[10001];
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for (int i = 0; i <= 10000; i++) {
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parent[i] = i;
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}
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}
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public int find(int x) {
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if (parent[x] != x) {
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parent[x] = find(parent[x]);
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}
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return parent[x];
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}
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public void union(int x, int y) {
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parent[find(x)] = find(y);
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}
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}
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}
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}

‎src/test/java/com/fishercoder/_721Test.java

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package com.fishercoder;
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import com.fishercoder.solutions._721;
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import org.junit.BeforeClass;
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import org.junit.Test;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.List;
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public class _721Test {
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private static _721.Solution1 solution1;
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private static _721.Solution2 solution2;
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private static List<List<String>> accounts;
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private static List<List<String>> expected;
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@BeforeClass
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public static void setup() {
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solution1 = new _721.Solution1();
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solution2 = new _721.Solution2();
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}
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@Test
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public void test1() throws Exception {
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accounts = new ArrayList<>();
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List<String> account1 = new ArrayList<>(Arrays.asList("John", "johnsmith@mail.com", "john00@mail.com"));
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List<String> account2 = new ArrayList<>(Arrays.asList("John", "johnnybravo@mail.com"));
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List<String> account3 = new ArrayList<>(Arrays.asList("John", "johnsmith@mail.com", "john_newyork@mail.com"));
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List<String> account4 = new ArrayList<>(Arrays.asList("Mary", "mary@mail.com"));
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accounts.add(account1);
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accounts.add(account2);
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accounts.add(account3);
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accounts.add(account4);
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expected = new ArrayList<>();
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List<String> expected1 = new ArrayList<>(Arrays.asList("Mary", "mary@mail.com"));
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List<String> expected2 = new ArrayList<>(Arrays.asList("John", "john00@mail.com", "john_newyork@mail.com", "johnsmith@mail.com"));
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List<String> expected3 = new ArrayList<>(Arrays.asList("John", "johnnybravo@mail.com"));
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expected.add(expected1);
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expected.add(expected2);
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expected.add(expected3);
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assertEqualsIgnoreOrdering(expected, solution1.accountsMerge(accounts));
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assertEqualsIgnoreOrdering(expected, solution2.accountsMerge(accounts));
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}
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private void assertEqualsIgnoreOrdering(List<List<String>> expected, List<List<String>> actual) throws Exception {
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//TODO: implement this method
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if (true) {
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return;
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} else {
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throw new Exception();
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}
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}
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@Test
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public void test2() throws Exception {
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accounts = new ArrayList<>();
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List<String> account1 = new ArrayList<>(Arrays.asList("Alex", "Alex5@m.co", "Alex4@m.co", "Alex0@m.co"));
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List<String> account2 = new ArrayList<>(Arrays.asList("Ethan", "Ethan3@m.co", "Ethan3@m.co", "Ethan0@m.co"));
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List<String> account3 = new ArrayList<>(Arrays.asList("Kevin", "Kevin4@m.co", "Kevin2@m.co", "Kevin2@m.co"));
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List<String> account4 = new ArrayList<>(Arrays.asList("Gabe", "Gabe0@m.co", "Gabe3@m.co", "Gabe2@m.co"));
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List<String> account5 = new ArrayList<>(Arrays.asList("Gabe", "Gabe3@m.co", "Gabe4@m.co", "Gabe2@m.co"));
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accounts.add(account1);
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accounts.add(account2);
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accounts.add(account3);
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accounts.add(account4);
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accounts.add(account5);
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expected = new ArrayList<>();
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List<String> expected1 = new ArrayList<>(Arrays.asList("Alex", "Alex0@m.co", "Alex4@m.co", "Alex5@m.co"));
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List<String> expected2 = new ArrayList<>(Arrays.asList("Kevin", "Kevin2@m.co", "Kevin4@m.co"));
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List<String> expected3 = new ArrayList<>(Arrays.asList("Ethan", "Ethan0@m.co", "Ethan3@m.co"));
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List<String> expected4 = new ArrayList<>(Arrays.asList("Gabe", "Gabe0@m.co", "Gabe2@m.co", "Gabe3@m.co", "Gabe4@m.co"));
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expected.add(expected1);
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expected.add(expected2);
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expected.add(expected3);
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expected.add(expected4);
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assertEqualsIgnoreOrdering(expected, solution1.accountsMerge(accounts));
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assertEqualsIgnoreOrdering(expected, solution2.accountsMerge(accounts));
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}
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@Test
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public void test3() throws Exception {
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accounts = new ArrayList<>();
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List<String> account1 = new ArrayList<>(Arrays.asList("David", "David0@m.co", "David1@m.co"));
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List<String> account2 = new ArrayList<>(Arrays.asList("David", "David3@m.co", "David4@m.co"));
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List<String> account3 = new ArrayList<>(Arrays.asList("David", "David4@m.co", "David5@m.co"));
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List<String> account4 = new ArrayList<>(Arrays.asList("David", "David2@m.co", "David3@m.co"));
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List<String> account5 = new ArrayList<>(Arrays.asList("David", "David1@m.co", "David2@m.co"));
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accounts.add(account1);
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accounts.add(account2);
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accounts.add(account3);
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accounts.add(account4);
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accounts.add(account5);
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expected = new ArrayList<>();
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List<String> expected1 = new ArrayList<>(Arrays.asList("David", "David0@m.co", "David1@m.co", "David2@m.co", "David3@m.co", "David4@m.co", "David5@m.co"));
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expected.add(expected1);
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assertEqualsIgnoreOrdering(expected, solution1.accountsMerge(accounts));
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assertEqualsIgnoreOrdering(expected, solution2.accountsMerge(accounts));
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}
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}

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