Isomorphic strings

Time: O(N); Space: O(1); easy

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.

[4]:

class Solution1(object):
    """
    Time: O(N)
    Space: O(1)
    """
    def isIsomorphic(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        if len(s) != len(t):
            return False

        s2t, t2s = {}, {}
        for p, w in zip(s, t):
            if w not in s2t and p not in t2s:
                s2t[w] = p
                t2s[p] = w
            elif w not in s2t or s2t[w] != p:
                # Contradict mapping.
                return False
        return True

[5]:

sol = Solution1()

s = "egg"
t = "add"
assert sol.isIsomorphic(s, t) == True

s = "foo"
t = "bar"
assert sol.isIsomorphic(s, t) == False

s = "paper"
t = "title"
assert sol.isIsomorphic(s, t) == True

[6]:

class Solution2(object):
    def isIsomorphic(self, s, t):
        if len(s) != len(t):
            return False

        return self.halfIsom(s, t) and self.halfIsom(t, s)

    def halfIsom(self, s, t):
        lookup = {}
        for i in range(len(s)):
            if s[i] not in lookup:
                lookup[s[i]] = t[i]
            elif lookup[s[i]] != t[i]:
                return False
        return True

[7]:

sol = Solution2()

s = "egg"
t = "add"
assert sol.isIsomorphic(s, t) == True

s = "foo"
t = "bar"
assert sol.isIsomorphic(s, t) == False

s = "paper"
t = "title"
assert sol.isIsomorphic(s, t) == True