Group Anagrams - Leetcode 49 - Hashmaps & Sets (Python)

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[GregHogg]

Master Data Structures & Algorithms For FREE at AlgoMap.io!

[swiftlyr]

Watched neetcode and some other videos on this problem and they sped through all of this code without properly explaining why we are doing what were doing. Happy to have found this video. Really solid explanation. Keep it up

[moezzzz]

just found this channel. your amazing man.

[teostorm]

slight note. return statement should be since the question asks to return a list of a list and not dict_values. great solution anyways

[anandreddy]

I've done like this
class Solution:
def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
counter = defaultdict(list)
for s in strs:
key = ''.join(sorted(s))
counter[key].append(s)
return list(counter.values())

[jmgpqcx]

fantastic example to help learner to realize that dict and sets are not hashable, since almost everyone would have thought of using a set of sets (i did) or dict with keys as dict (like you did) and realize that it's not possible

[benjaminappiah]

I love the solution. I am curious as to why using dict is faster than defaultdict

[ariellockett]

im confused about if the key isn't in the dictionary yet, it associates its value with an empty list. how does the first string in a list get added when the key isn't in the dictionary yet if it associates an empty list and not the appended string? hope this makes sense

[matveysaprykin]

What’s the problem converting dict to str and pass it as a key?

[FUNTasticFlutter]

my solution ..runtime 87 19.3 MB
def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
anagram_dict={}
for s in strs:
sorted_s= "".join(sorted(s))
if sorted_s in anagram_dict:

else:
anagram_dict[sorted_s]=[s]

return list(anagram_dict.values())

[Ak-ggs]

4:58 Note : ord is used to get the unicode of a charecter and not ascii value

[hlubradio]

I don't think I'd need to learn this one ... lol