class Solution(object):

def isAnagram(self, s, t):

"""

:type s: str

:type t: str

:rtype: bool

"""

"""

# --- List comparison method: Time Limit Exceeded

# Anagram validity check

if len(s) > len(t):

check = list(s)

other = list(t)

else:

check = list(t)

other = list(s)

for char in check:

if char not in other:

return False

else:

del other[other.index(char)]

return True

# What if unicode characters occur, solution?

"""

"""

# 100pass = To check if same number of elements occur in both the strings

# Python Hashmap method using collections counter

sc = collections.Counter(s)

tc = collections.Counter(t)

if sc == tc:

return True

else:

return False

"""

"""

# 100 pass - using manual naive hashmap

sc = {}

tc = {}

for char in s:

if char not in sc:

sc[char] = 0

sc[char] += 1

for char in t:

if char not in tc:

tc[char] = 0

tc[char] += 1

if sc == tc:

return True

else:

return False

"""

# 100pass = Easy pythonic hack - using sorted both the sides

if sorted(s) == sorted(t):

return True

else:

return False