Question

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Given two (dictionary) words as Strings, determine if they are isomorphic.

Two words are called isomorphic if the letters in one word can be remapped to get the second word. Remapping a letter means replacing all occurrences of it with another letter while the ordering of the letters remains unchanged. No two letters may map to the same letter, but a letter may map to itself.

Example:

given "foo", "app"; returns true
we can map 'f' -> 'a' and 'o' -> 'p'

given "bar", "foo"; returns false
we can't map both 'a' and 'r' to 'o'

given "turtle", "tletur"; returns true
we can map 't' -> 't', 'u' -> 'l', 'r' -> 'e', 'l' -> 'u', 'e' -'r'

given "ab", "ca"; returns true
we can map 'a' -> 'c', 'b'

Solution

My first thought was: map char to char, then check in the hashmap. However, this will not work!

input: abc, pzz
check a -> p
check b -> z
check c -> z

Using a hashmap ** does not show us **whether 2 chars match to the same char__. In above example, b and c both match to z.

Best Solution, suggested by urik:

HashMap (char, firstSeenIndice) for each string. The encoding of firstSeenIndices shud match.

E.g. Foo and app both encode to 011

Abcd and hole both encode to 0123

Code

public boolean isomorphic(String s, String t) {
    if (s.length() != t.length()) {
        return false;
    }
    return (sequence(s).equals(sequence(t)));
}

private String sequence(String s) {
    HashMap<Character, Integer> map = new HashMap<Character, Integer>();
    StringBuilder sb = new StringBuilder();

    for (int i = 0; i < s.length(); i++) {
        if (map.containsKey(s.charAt(i))) {
            sb.append(map.get(s.charAt(i)));
        } else {
            map.put(s.charAt(i), i);
        }
    }
    return sb.toString();
}