# Time: O(N · M)
# Space: O(N)

from collections import Counter, defaultdict

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        # If key doesn't exist, creates empty list using `list`
        # factory/ctor function
        groups = defaultdict(list)
        
        for s in strs:
            # Keep counter for num of times each letter appeared
            counts = [0] * 26
            
            for c in s:
                i = ord(c) - ord('a')
                counts[i] += 1
            
            # Lists aren't hashable, so we need to convert the counts
            # to tuple so that we can key them
            groups[tuple(counts)].append(s)
        
        return list(groups.values())