Explain the use of a bit vector for determining if all characters are unique

Question

I am confused about how a bit vector would work to do this (not too familiar with bit vectors). Here is the code given. Could someone please walk me through this?

Answer 1

Lets break down the code line by line.

int checker = 0; We are initiating a checker which will help us find duplicate values.

int val = str.charAt(i) - 'a'; We are getting the ASCII value of the character at the 'i'th position of the string and subtracting it with the ASCII value of 'a'. Since the assumption is that the string is lower characters only, the number of characters in limited to 26. Hece, the value of 'val' will always be >= 0.

if ((checker & (1 << val)) > 0) return false;

checker |= (1 << val);

Now this is the tricky part. Lets us consider an example with string "abcda". This should ideally return false.

For loop iteration 1:

Checker: 00000000000000000000000000000000

val: 97-97 = 0

1 << 0: 00000000000000000000000000000001

checker & (1 << val): 00000000000000000000000000000000 is not > 0

Hence checker: 00000000000000000000000000000001

For loop iteration 2:

Checker: 00000000000000000000000000000001

val: 98-97 = 1

1 << 0: 00000000000000000000000000000010

checker & (1 << val): 00000000000000000000000000000000 is not > 0

Hence checker: 00000000000000000000000000000011

For loop iteration 3:

Checker: 00000000000000000000000000000011

val: 99-97 = 0

1 << 0: 00000000000000000000000000000100

checker & (1 << val): 00000000000000000000000000000000 is not > 0

Hence checker: 00000000000000000000000000000111

For loop iteration 4:

Checker: 00000000000000000000000000000111

val: 100-97 = 0

1 << 0: 00000000000000000000000000001000

checker & (1 << val): 00000000000000000000000000000000 is not > 0

Hence checker: 00000000000000000000000000001111

For loop iteration 5:

Checker: 00000000000000000000000000001111

val: 97-97 = 0

1 << 0: 00000000000000000000000000000001

checker & (1 << val): 00000000000000000000000000000001 is > 0

Hence return false.