3 problems completed

[BharathVuppala96]

No description provided.

[super30admin]

Your solution for grouping anagrams is correct and demonstrates a good understanding of the problem. Here are some suggestions for improvement:

• Consider using a different variable name instead of map, such as anagram_groups or groups, to avoid shadowing the built-in map function.

• While using the sorted string as a key is acceptable, for longer strings, you might consider using a tuple of character counts for better performance. For example:

  key = tuple(sorted(s))   # This is similar to what you have, but as a tuple which is hashable.
  

  Alternatively, you can use a frequency array converted to a tuple:

  count = [0] * 26
  for char in s:
      count[ord(char) - ord('a')] += 1
  key = tuple(count)
  

  This method has a time complexity of O(N * K) which is better when K is large.

• You can use defaultdict from the collections module to make the code more concise:

  from collections import defaultdict
  groups = defaultdict(list)
  for s in strs:
      key = ''.join(sorted(s))
      groups[key].append(s)
  return list(groups.values())
  

  This avoids the need to check if the key exists.

Overall, your solution is correct and efficient for most cases. Well done!