# 49. Group anagrams

https://leetcode.com/problems/group-anagrams/

Given an array of strings strs, group the anagrams together. You can return the answer in any order. An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1: Input: strs = [“eat”,“tea”,“tan”,“ate”,“nat”,“bat”] Output: [[“bat”],[“nat”,“tan”],[“ate”,“eat”,“tea”]] Example 2: Input: strs = [""] Output: [[""]] Example 3: Input: strs = [“a”] Output: [[“a”]]

Constraints:

``````1 <= strs.length <= 104
0 <= strs[i].length <= 100
strs[i] consists of lowercase English letters.
``````

• code
``````class Solution {

public String sortString(String word){
char[] l = word.toCharArray();
Arrays.sort(l);
return String.valueOf(l);
// return Stream.of(word.split("")).sorted().collect(Collectors.joining());
}

public List<List<String>> groupAnagrams(String[] strs) {
Map<String, List> m = new HashMap<>();
for (String word: strs){
String sorted = sortString(word);
if (!m.containsKey(sorted)) m.put(sorted, new ArrayList());
}
return new ArrayList(m.values());
}
}
``````
• code notice after sorted “eat”, it becomes [‘a’,‘e’,’t']
``````class Solution:
def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
dic_ana = {}
for word in strs:
sword = "".join(sorted(word))
if sword not in dic_ana:
dic_ana[sword] = [word]
else:
dic_ana[sword].append(word)

return [v for v in dic_ana.values()]

``````