Question

I need to check in Java if a word consists of unique letters (case insensitive). As straight solution is boring, I came up with:

1. For every char in a string check if indexOf(char) == lastIndexOf(char).
2. Add all chars to HashSet and check if set size == string length.
3. Convert a string to a char array, sort it alphabetically, loop through array elements and check if c[i] == c[i+1].

Currently I like #2 the most, seems like the easiest way. Any other interesting solutions?

Answer 1

I don’t like 1. — it’s an O(N²) algorithm. Your 2. is roughly linear, but always traverses the entire string. Your 3. is O(N lg₂ N), with (probably) a relatively high constant — probably almost always slower than 2.

My preference, however, would be when you try to insert a letter into the set, check whether it was already present, and if it was, you can stop immediately. Given random distribution of letters, this should require scanning only half the string on average.

Edit: both comments are correct that exactly what portion of the string you expect to scan will depend on the distribution and the length — at some point the string is long enough that a repeat is inevitable, and (for example) one character short of that, the chance is still pretty darned high. In fact, given a flat random distribution (i.e., all characters in the set are equally likely), this should fit closely with the birthday paradox, meaning the chance of a collision is related to the square root of the number of possible characters in the character set. Just for example, if we assumed basic US-ASCII (128 characters) with equal probability, we’d reach a 50% chance of a collision at around 14 characters. Of course, in real strings we could probably expect it sooner than that, since the ASCII characters aren’t used with anywhere close to equal frequency in most strings.