# Question

Given two strings s and t, determine if they are isomorphic.

Two strings are isomorphic if the characters in s can be replaced to get t.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.

Example 1:

Input: s = "egg", t = "add"
Output: true


Example 2:

Input: s = "foo", t = "bar"
Output: false


Example 3:

Input: s = "paper", t = "title"
Output: true


Note:

You may assume both s and t have the same length.

# Solution

class Solution(object):
def isIsomorphic(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
if len(s) != len(t):
return False

checked = [0] * len(s)
for i in range(len(s)):
if checked[i] == 0:
checked[i] = 1
for j in range(i + 1, len(s)):
if s[j] == s[i]:
if t[j] == t[i]:
checked[j] = 1
else:
return False
elif t[j] == t[i]:
return False

for c in checked:
if c == 0:
return False

return True



# Other Solutions

The basic idea is that occurrences of same character should be the same.

def isIsomorphic1(self, s, t):
d1, d2 = {}, {}
for i, val in enumerate(s):
d1[val] = d1.get(val, []) + [i]
for i, val in enumerate(t):
d2[val] = d2.get(val, []) + [i]
return sorted(d1.values()) == sorted(d2.values())

def isIsomorphic2(self, s, t):
d1, d2 = [[] for _ in xrange(256)], [[] for _ in xrange(256)]
for i, val in enumerate(s):
d1[ord(val)].append(i)
for i, val in enumerate(t):
d2[ord(val)].append(i)
return sorted(d1) == sorted(d2)

def isIsomorphic3(self, s, t):
return len(set(zip(s, t))) == len(set(s)) == len(set(t))

def isIsomorphic4(self, s, t):
return [s.find(i) for i in s] == [t.find(j) for j in t]

def isIsomorphic5(self, s, t):
return list(map(s.find, s)) == list(map(t.find, t))

def isIsomorphic(self, s, t):
d1, d2 = [0 for _ in xrange(256)], [0 for _ in xrange(256)]
for i in xrange(len(s)):
if d1[ord(s[i])] != d2[ord(t[i])]:
return False
d1[ord(s[i])] = i+1
d2[ord(t[i])] = i+1
return True