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  3. Strings
  4. Java Anagrams
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Java Anagrams

  • vky8411
     + 5 comments

    Great explanation!

    But I believe you can still optimize it a little bit. How about using char array instead of charAt() function? Because every call to charAt() method contains internall check for ArrayOutOfBoundException.

    My vision:

    static boolean isAnagram(String a, String b) {
        if ((a == null || b == null) || (a.length() != b.length())) {
            return false;
        }
        final char[] ARRAY_A = a.toUpperCase().toCharArray();
        final Map<Character, Integer> map = new HashMap<>();
        for (int i = 0; i < ARRAY_A.length; i ++) {
            if (map.containsKey(ARRAY_A[i])) {
                map.put(ARRAY_A[i], (map.get(ARRAY_A[i]) + 1));
            } else {
                map.put(ARRAY_A[i], 1);
            }
        }
        final char[] ARRAY_B = b.toUpperCase().toCharArray();
        for (int i = 0; i < ARRAY_A.length; i ++) {
            if (map.containsKey(ARRAY_B[i])) {
                Integer count = map.get(ARRAY_B[i]);
                if (count == 0) {
                    return false;
                } else {
                    count --;
                    map.put(ARRAY_B[i], count);
                }
            } else {
                return false;
            }
        }
        return true;
    }