package com.freetymekiyan.algorithms.level.medium;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
/**
* Given an array of strings, group anagrams together.
* <p>
* For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"],
* Return:
* <p>
* | [
* | ["ate", "eat","tea"],
* | ["nat","tan"],
* | ["bat"]
* | ]
* Note: All inputs will be in lower-case.
* <p>
* Company Tags: Amazon, Bloomberg, Uber, Facebook, Yelp
* Tags: Hash Table, String
* Similar Problems: (E) Valid Anagram, (E) Group Shifted Strings
*/
public class GroupAnagrams {
/**
* Hash Table.
* Use sorted word as the key, since all anagrams share the same sorted form.
* Then same anagrams form a list as the value.
* For each word, sort its character array and build a string key.
* If the key is not in map yet, add an empty array list to map.
* Then add the word to the list.
* <p>
* A possible improvement is sorting the word using counting sort. Two pass.
* Since all inputs are lowercase characters.
* Reduce sorting time from O(nlogn) to O(n).
*/
public List<List<String>> groupAnagrams(String[] strs) {
if (strs == null || strs.length == 0) {
return Collections.emptyList();
}
Map<String, List<String>> map = new HashMap<>();
for (int i = 0; i < strs.length; i++) {
char[] word = strs[i].toCharArray();
Arrays.sort(word);
String key = String.valueOf(word);
if (!map.containsKey(key)) {
map.put(key, new ArrayList<>());
}
map.get(key).add(strs[i]);
}
return new ArrayList<>(map.values());
}
}