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validAnagram.py
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79 lines (62 loc) · 1.63 KB
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# https://leetcode.com/problems/valid-anagram/?envType=study-plan&id=data-structure-i
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
# ss = sorted(s)
# tt = sorted(t)
# return ss==tt
# OR
# if len(s)!=len(t):
# return False
# for i in s:
# if i not in t:
# return False
# else:
# d = {}
# d1 = {}
# for j in s:
# if j in d:
# d[j]+=1
# else:
# d[j] = 1
# for k in t:
# if k in d1:
# d1[k]+=1
# else:
# d1[k] = 1
# for l in d:
# if d[l]!=d1[l]:
# return False
# return True
# OR
# l = []
# for i in t:
# l.append(i)
# for j in s:
# if j not in l:
# return False
# l.remove(j)
# if l:
# return False
# else:
# return True
# OR
if len(s)!=len(t):
return False
sd ,td = defaultdict(int),defaultdict(int)
for i in s:
sd[i]+=1
for j in t:
td[j]+=1
for k in s:
if sd[k]!=td[k]:
return False
else:
return True
# OR
# if len(s)!=len(t): return False
# s = sorted(s)
# t = sorted(t)
# for i in range(len(s)):
# if s[i]!=t[i]:
# return False
# return True