LeetCode 49

49. Group Anagrams

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Problem

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.

Solution(Sorting based)

from collections import defaultdict

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        anagrams = collections.defaultdict(list)

        for word in strs:
            anagrams[''.join(sorted(word))].append(word)
        return anagrams.values()

Time and Space Complexity

• Sorting a word: O(k log k), where k is the length of the word
• Total time complexity: O(n * k log k), where n is the number of words in the input list
• Space complexity: O(n * k) for storing the grouped anagrams in a hash map

Solution(word length count based)

from collections import defaultdict

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        anagrams = collections.defaultdict(list)

        for word in strs:
            count = [0] * 26 # a~z
            for chat in word:
                count[ord(char) - ord('a')] += 1

            key = tuple(count) # List can't be keys, so convert to tuples
            anagrams[key].append(word)

Time and Space Complexity

• Counting letters in a word takes O(k), where k is the length of the word

• Total time complexity is O(n * k), where n is the number of words in the input list

• Space complexity is O(n * k) for storing the grouped anagrams in a hash map (defaultdict)