Detecting if a string has unique characters: comparing my solution to "Cracking the Coding Interview?"
Solution 1:
There are two separate questions here: what's the efficiency of your solution, and what is the reference solution doing? Let's treat each independently.
First, your solution:
public static boolean checkForUnique(String str){
boolean containsUnique = false;
for(char c : str.toCharArray()){
if(str.indexOf(c) == str.lastIndexOf(c)){
containsUnique = true;
} else {
containsUnique = false;
}
}
return containsUnique;
}
Your solution essentially consists of a loop over all characters in the string (let's say there are n of them), checking on each iteration whether the first and last index of the characters are the same. The indexOf and lastIndexOf methods each take time O(n), because they have to scan across all the characters of the string to determine if any of them match the one you're looking for. Therefore, since your loop runs O(n) times and does O(n) work per iteration, its runtime is O(n2).
However, there's something iffy about your code. Try running it on the string aab. Does it work correctly on this input? As a hint, as soon as you determine that there are two or more duplicated characters, you're guaranteed that there are duplicates and you can return that not all characters are unique.
Now, let's look at the reference:
public static boolean isUniqueChars(String str) {
if (str.length() > 256) { // NOTE: Are you sure this isn't 26?
return false;
}
int checker = 0;
for (int i = 0; i < str.length(); i++) {
int val = str.charAt(i) - 'a';
if ((checker & (1 << val)) > 0) return false;
checker |= (1 << val);
}
return true;
}
This solution is cute. The basic idea is the following: imagine that you have an array of 26 booleans, each one tracking whether a particular character has appeared in the string already. You start with all of them false. You then iterate across the characters of the string, and each time you see a character you look into the array slot for that character. If it's false, this is the first time you've seen the character and you can set the slot to true. If it's true, you've already seen this character and you can immediately report that there's a duplicate.
Notice that this method doesn't allocate an array of booleans. Instead, it opts for a clever trick. Since there are only 26 different characters possible and there are 32 bits in an int, the solution creates an int variable where each bit of the variable corresponds to one of the characters in the string. Instead of reading and writing an array, the solution reads and writes the bits of the number.
For example, look at this line:
if ((checker & (1 << val)) > 0) return false;
What does checker & (1 << val) do? Well, 1 << val creates an int value that has all bits zero except for the valth bit. It then uses bitwise AND to AND this value with checker. If the bit at position val in checker is already set, then this evaluates to a nonzero value (meaning we've already seen the number) and we can return false. Otherwise, it evaluates to 0, and we haven't seen the number.
The next line is this:
checker |= (1 << val);
This uses the "bitwise OR with assignment" operator, which is equivalent to
checker = checker | (1 << val);
This ORs checker with a value that has a 1 bit set only at position val, which turns the bit on. It's equivalent to setting the valth bit of the number to 1.
This approach is much faster than yours. First, since the function starts off by checking if the string has length greater than 26 (I'm assuming the 256 is a typo), the function never has to test any string of length 27 or greater. Therefore, the inner loop runs at most 26 times. Each iteration does O(1) work in bitwise operations, so the overall work done is O(1) (O(1) iterations times O(1) work per iteration), which is significantly faster than your implementation.
If you haven't seen bitwise operations used this way, I'd recommend searching for "bitwise operators" on Google to learn more.
Hope this helps!
Solution 2:
The book solution is one I don't like and I believe is dysfunctional..... templatetypedef has posted a comprehensive answer which indicates that the solution is a good one. I disagree, since the book's answer assumes that the string only has lower-case characters, (ascii) and does no validation to ensure that.
public static boolean isUniqueChars(String str) {
// short circuit - supposed to imply that
// there are no more than 256 different characters.
// this is broken, because in Java, char's are Unicode,
// and 2-byte values so there are 32768 values
// (or so - technically not all 32768 are valid chars)
if (str.length() > 256) {
return false;
}
// checker is used as a bitmap to indicate which characters
// have been seen already
int checker = 0;
for (int i = 0; i < str.length(); i++) {
// set val to be the difference between the char at i and 'a'
// unicode 'a' is 97
// if you have an upper-case letter e.g. 'A' you will get a
// negative 'val' which is illegal
int val = str.charAt(i) - 'a';
// if this lowercase letter has been seen before, then
// the corresponding bit in checker will have been set and
// we can exit immediately.
if ((checker & (1 << val)) > 0) return false;
// set the bit to indicate we have now seen the letter.
checker |= (1 << val);
}
// none of the characters has been seen more than once.
return true;
}
The bottom line, given templatedef's answer too, is that there's not actually enough information to determine whether the book's answer is right.
I distrust it though.
templatedef's answer on the complexity is one I agree with though.... ;-)
EDIT: As an exercise, I converted the book's answer to one that will work (albeit slower than the book's answer - BigInteger is slow).... This version does the same logic as the book's, but does not have the same validation and assumption problems (but it is slower). It is useful to show the logic too.
public static boolean isUniqueChars(String str) {
if (str.length() > 32768) {
return false;
}
BigInteger checker = new BigInteger(0);
for (int i = 0; i < str.length(); i++) {
int val = str.charAt(i);
if (checker.testBit(val)) return false;
checker = checker.setBit(val);
}
// none of the characters has been seen more than once.
return true;
}