Valid Anagram - Leetcode 242 - Hashmaps & Sets (Python)

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[GregHogg]

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

[hamzaunsal]

Another Solution :
def isAnagram(self, s, t):
return sorted(s) == sorted(t)

[MrIrishBoss]

I used a similar approach to what I did in Ransom Note.
for i in set(s):
if s.count(i) != t.count(i) or len(s) != len(t):
return False
return True

[TheTrackoShow]

please do not sort or use Counter, if you are serious about getting a job, this is the same code from the explain but if not the most optimal since we are not checking for unicode values which can be done by using ord()



class Solution:
def isAnagram(self, s: str, t: str) -> bool:
if len(s) != len(t): # return false if the are not the same length
return False

countS, countT = {}, {}

for index in range(len(s)):
countS[s[index]] = countS.get(s[index], 0) + 1 # get the value if it does not exist set it to 0, else add 1
countT[t[index]] = countT.get(t[index], 0) + 1 # get the value if it does not exist set it to 0, else add 1

return countS == countT # return true or false if the are equal

[dileepkonchada]

Since the anagrams are in lowercase alphabets
So, if we check the frequency of characters with an array of size 26
We iterate through array
One anagram is to count the frequency and the other angaram used to decrease the frequency after that
In the end if we found any element in array Greater thank 0 return false otherwise return true
This approach also works in O(n) time complexity and with constant space complexity

[CharlyBraga]

Is the time complexity really O(n)? I ask it because the Counter need run through list to creat the dictionary and sort the last one.

[christianjt]

nice short solution, thanks for sharing

[helloworldcsofficial]

Do you recommend using Counter in an actual interview?

[jorgesanabria]

This is an awesome series keep it up 👍🏾

[RealDarkFilm]

With all due respect, this is a horrible solution for a coding interview. Never import during leetcode interview. They dont want you taking the easy way out