public boolean evaluate2(String input) {
    validateInput(input);

    int[] chars = new int[256];
    for (char c : input.toCharArray()) {
      if (chars[c] >= 1) {
        return false;
      } else {
        chars[c]++;
      }
    }
    return true;
  }
Faster solution to this problem. This solution is based on one important detail, we are
assuming ASCII as charset. This solution is also iterative and changes the HashMap
implementation with an array with a size equivalent to the maximum different chars you can
find in ASCII. The complexity order of this algorithm in time terms is O(N), like the previous
algorithm. In space terms, the complexity order of this algorithm is O(1).
src/main/java/com/github/pedrovgs/problem50/UniqueChars.java