BitSet bug in cracking the coding interview?

Question

The following is the implementation of BitSet in the solution of question 10-4 in cracking the coding interview book. Why is it allocating an array of size/32 not (size/32 + 1). Am I missing something here or this is a bug?

If I pass 33 to the constructor of BitSet then I will allocate only one int and If I try to set or get the bit 32, I will get an AV!

package Question10_4;

class BitSet {
    int[] bitset;

    public BitSet(int size) {
            bitset = new int[size >> 5]; // divide by 32
    }

    boolean get(int pos) {
            int wordNumber = (pos >> 5); // divide by 32
            int bitNumber = (pos & 0x1F); // mod 32
            return (bitset[wordNumber] & (1 << bitNumber)) != 0;
    }

    void set(int pos) {
            int wordNumber = (pos >> 5); // divide by 32
            int bitNumber = (pos & 0x1F); // mod 32
            bitset[wordNumber] |= 1 << bitNumber;
    }

}

Answer 1

Yes, answer in the book is incorrect. Correct answer:

bitset = new int[(size + 31) >> 5]; // divide by 32

Answer 2

From what I can gather from reading the solution you mention (on page 205), and the little I understand about computer programming, it seems to me that this is a special implementation of a bitset, meant to take the argument of 32,000 in its construction (see the checkDuplicates function. The question is about examining an array with numbers from 1 to N, where N is at most 32,000, with only 4KB of memory).

This way, an array of 1000 elements is created, each one used for 32 bits in the bit set. You can see in the bitset class that to get a bit's position, we (floor) divide by 32 to get the array index, and then mod 32 to get the specific bit position.