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Mahmoud EL-kariouny
Mahmoud EL-kariouny

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Python challenge_7


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.


  • 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
            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
            str_dic_tow[str_tow] = 1

    if str_dic_one == str_dic_tow:
        return True
        return False

print(is_anagram('typhoon', 'opython'))
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Another solution
def is_anagram(string1, string2):
    return sorted(string1) == sorted(string2)
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def count_letters(string):
    return {l: string.count(l) for l in string}
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def is_anagram(string1, string2):
    return count_letters(string1) == count_letters(string2)
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Add your solution in the comment :)

Discussion (1)

totally_chase profile image
Phantz • Edited on
from collections import Counter

is_anagram = lambda x, y: Counter(x) == Counter(y)
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I'm gonna guess Counter would be able to do it pretty fast, if not the fastest.