Day042 - 49. Group Anagrams

49. Group Anagrams

Given an array of strings strs, group the

anagrams

together. You can return the answer in any order.

Example 1:

Input: strs = [“eat”,”tea”,”tan”,”ate”,”nat”,”bat”] Output: [[“bat”],[“nat”,”tan”],[“ate”,”eat”,”tea”]] Explanation:

• There is no string in strs that can be rearranged to form "bat".
• The strings "nat" and "tan" are anagrams as they can be rearranged to form each other.
• The strings "ate", "eat", and "tea" are anagrams as they can be rearranged to form each other.

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.

내 풀이

from collections import Counter, defaultdict

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        """
        strs = ["eat","tea","tan","ate","nat","bat"]
        """

        # checker = defaultdict(list)

        # for idx, i in enumerate(strs):
        #     counter = sorted(Counter(i).items(), key=lambda x: x[0])
        #     checker[''.join([j[0] + str(j[1]) for j in counter])].append(idx)
            

        sorted_strs = [str(''.join(sorted(i))) for i in strs]

        checker = defaultdict(list)

        for idx, i in enumerate(sorted_strs):
            checker[i].append(idx)

        answer = []
        for i in checker.values():
            temp_list = []
            for j in i:
                temp_list.append(strs[j])
            answer.append(temp_list)
                

        return answer

# Time Complexity : \(O(N)\)