[ASSIGNMENT] String Anagram.

Question

Write a program to check if two given String is Anagram of each other.
Your program should print true if two Strings are Anagram, false otherwise.
A string is said to be an anagram if it contains same characters and same length but in different order.

For example:
If the input strings are 'listen' and 'silent'.
The output should be True.

P.S. I submit the challenges in lesson factory so that they can be made assignments.But till they get approved i post them here so as to help everyone practice.

Answer 1

I guess I was too insensitive 😂🤣🤣🤣🤣🤣🤣
updated :-
print(sorted(input().lower()) == sorted(input().lower()))

Answer 2

https://code.sololearn.com/cHkV3NUt860r/?ref=app

Answer 3

#My try in Python (1Liner) :-
print(sorted(input())==sorted(input()))

edit :- check the updated version...

Answer 4

https://code.sololearn.com/csPALI7Bsj9S/?ref=app
https://code.sololearn.com/c69l7OOlCw04/?ref=app

Answer 5

https://code.sololearn.com/czMBi9u62O6t/?ref=app

Answer 6

@mitali jadhavrao
Thank you 😉

Answer 7

# python

def are_anagrams(s, ss):
a = sorted(s.lower())
b = sorted(ss.lower())
return a == b

# EDIT: lists not needed, indeed ☺️

Answer 8

Hechmi Barhouma Thank you for pointing it out.👍👍
Also ur solution is great.
Thanks for participating in this challenge. 😊

Answer 9

Pedro Demingos Good you have also taken care of the case.👍

Answer 10

Hechmi Barhouma It would check only the first matching character n return True.It wont go to next character because of return statement.
Make few changes.
Nice try👍

Answer 11

Hechmi Barhouma:
In addition to the problem pointed by mitali jadhavrao, you could have false positives with your logical mistake...
Testing if each letter of a word is present in the other one is not enough to proove that they're anagram ^^ You need also test that other one don't have letter not used in first word, and that each word use same count of each letter ;P

Python short solution:

def is_anagram(a,b):
return ''.join(sorted(list(a))) == ''.join(sorted(list(b)))

print(is_anagram('silent','listen')) # True

Answer 12

🌛DT🌜 Great logic.👍👍
Thanks for participating.

Answer 13

Keegorp Good try.👍👍
The code is awesome.
Thanks for taking part😊

Answer 14

hinanawi Both the codes are great.👍

Answer 15

LukArToDo Good one 👍

Answer 16

a ruby one

https://code.sololearn.com/ctGugd2gk7jS/?ref=app

Answer 17

🌛DT🌜:
Good point: I'm too much poluated by JS, and I've forgotten that sorted() doesn't require a list as argument, but an iterable such as a string, and that Python could compare list, without the need to join() them to compare two strings :P

Answer 18

Russ Well i dont see any reason why someone has downvoted.
Just a suggestion ...see for cases.
Nice try👍.

Answer 19

https://code.sololearn.com/cA4kA38tcNk1/?ref=app
https://code.sololearn.com/wi8UXmC9G0sN/?ref=app

Answer 20

My answer:
https://code.sololearn.com/cZhap0UB8JuW/#c