## DEV Community # Python challenge_7

### Anagrams

#### level of challenge 3/10

• Two strings are anagrams
• if you can make one from the other by rearranging the letters.
• Write a function
• named is_anagram that takes two strings as its parameters.
• Your function should return True if the strings are anagrams,
• and False otherwise.

• For example,

• the call is_anagram("typhoon", "opython") should return True while the call

• is_anagram("Alice", "Bob") should return False.

Hint

• You can compare how many times each letter appears in each string.
• Alternatively, sorting the letters in each string makes this much easier.
##### My solution
``````def is_anagram(string_one, string_two):
str_dic_one = {}
str_dic_tow = {}

for str_one in string_one:
if str_one in str_dic_one:
str_dic_one[str_one] += 1
else:
str_dic_one[str_one] = 1

for str_tow in string_two:
if str_tow in str_dic_tow:
str_dic_tow[str_tow] += 1
else:
str_dic_tow[str_tow] = 1

if str_dic_one == str_dic_tow:
return True
else:
return False

print(is_anagram('typhoon', 'opython'))
``````
##### Another solution
``````def is_anagram(string1, string2):
return sorted(string1) == sorted(string2)
``````
``````def count_letters(string):
return {l: string.count(l) for l in string}
``````
``````def is_anagram(string1, string2):
return count_letters(string1) == count_letters(string2)
`````` Thanh Tran
``````def is_anagram(word1,word2):
if sorted(word1)== sorted(word2):
return True
else:
return False
`````` Phantz • Edited
``````from collections import Counter

is_anagram = lambda x, y: Counter(x) == Counter(y)
``````

I'm gonna guess `Counter` would be able to do it pretty fast, if not the fastest.

DEV Community

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