O(n) python solution using hashmap and set

  • 0

    class Solution(object):

    def isIsomorphic(self, s, t):
        :type s: str
        :type t: str
        :rtype: bool
        history = {}
        used_chars = set()
        if len(s) != len(t):
            return False
        for i in range(len(s)):
            if s[i] not in history:
                if t[i] in used_chars:
                    return False
                history[s[i]] = t[i]
            elif history[s[i]] != t[i]:
                return False
        return True

    hashmap is used to keep history of which 's' char is assigned which 't' char and set is used to verify we are not assigning the same character twice.

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