We use a hash map where the key is either the sorted version of the string or a frequency tuple.

1. Initialize ans = defaultdict(list).
2. For each string s in strs:
   • Count character frequencies (a-z) into a tuple of size 26.
   • Use this tuple as a key in ans and append s to the list.
3. Return ans.values().

• Time Complexity: O(N * K), where K is the length of the string.
• Space Complexity: O(N * K).

from collections import defaultdict

def group_anagrams(strs):
    ans = defaultdict(list)
    for s in strs:
        count = [0] * 26
        for char in s:
            count[ord(char) - ord('a')] += 1
        ans[tuple(count)].append(s)
    return list(ans.values())