242. Valid Anagram leetcode.

Question

Given two strings s and t, return true if t is an anagram of s, and false otherwise.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

My code gives false when it should give true.

class Solution(object):
def isAnagram(self, s, t):
sed = set()
sed2 = set()
for el in s:
sed.add(el)
for e in t:
sed2.add(e)

for el in sed2:
if el not in sed:
print("false")
print("true")

Answer 1

You should print the result of the function.

print(isAnagram("abc", "cba"))

Answer 2

Paul-Henry Paltmann ,

calling the function with:

> "abcde", "beadx"

the output is:

false
true

>> a different (and simple) approach can be to sort both of the strings and then compare them for equalty.

Answer 3

Converting to set, keeps only unique letters. Duplicates are removed.
For example, ESSAY is not anagram with EASY, even though they have the same letters, but each letter is not used exactly once in the other word.

In your function you should return the result, rather than printing it. If there are multiple mismatches, "false" can be printed multiple times, then at the end, "true" is printed.

Answer 4

Paul-Henry Paltmann Two side hints:
1. You don't need a class for this. There won't be objects, or any data persistence during code execution. A simple function is enough.
2. Use meaningful variable names. They help in understanding the logic.

Answer 5

Emerson Prado I agree with you on both points, but actually these parts are given by the Leetcode solution template.

https://leetcode.com/problems/valid-anagram/

One good element of this scaffolding that was not followed though: the isAnagram function is not supposed to print anything, it is supposed to return a bool value (True or False).

Answer 6

The sort() method does an in-place sorting of the list, and does not return anything.
There is also a sorted() function which leaves the list in its original state and returns a new, sorted list.

So there are two ways you can fix it (apply to both lists):

list1.sort()
# and then compare the original lists

or

srtdlst1 = sorted(list1)

Read how to use sort and sorted:
https://realpython.com/JUMP_LINK__&&__python__&&__JUMP_LINK-sort/

Answer 7

Yes, thanks for your help

Answer 8

By the way you are now on the right track about finding the correct solution.
There is a lot that can be simplified.
The sorted function can actually take any iterable, that includes a string, and converts it automatically to a sorted list of characters.
So ultimately you can do the conversion and the comparison in a single step, without using intermediary variables.
The function should return the boolean result of the comparison, so even your if condition is not necessary.
Simplified version of your function looks like this:

def isAnagram(s, t):
return sorted(s) == sorted(t)

Answer 9

So this code still returns None.... I did it exactly like it should be

def isAnagram(s, t):
list1= []
list2 = []
list1.extend(s)
list2.extend(t)
srtdlst1= list1.sort()
srtdlst2= list2.sort()

if srtdlst1 == srtdlst2:
return True
else:
return False
isAnagram("anagram", "nagaram")

Answer 10

Tibor Santa the code still outputs nothing

https://www.sololearn.com/compiler-playground/cAiiQG5SSIPj

Answer 11

wtf