You have an array with all the numbers from 1 to N, where N is at most 32,000. The array may have duplicate entries and you do not know what N is. With only 4KB of memory available, how would you print all duplicate elements in the array?

**My initial thoughts:**

4KB = 32768 bits. We can use a byte array to represent if we have seen number i. If so, we make it 1. If we encounter a duplicate, we would have marked the particular position with 1, so we would know we have a duplicate. Then we print it out.

**My initial codes:**

public static void findDuplicates(int[] array) {
byte[] bitVector = new byte[32000 / 8];
for (int number : array) {
int posInArray = number >> 3;
int posInBit = number % 8;
if ((bitVector[posInArray] & (1 << posInBit)) > 0)
// found duplicates
System.out.println(number);
else
bitVector[posInArray] |= (1 << posInBit);
}
}

**Solution:**

We have 4KB of memory which means we can address up to 8 * 4 * (2^10) bits. Note that 32*(2^10) bits is greater than 32000. We can create a bit vector with 32000 bits, where each bit represents one integer.

NOTE: While this isn’t an especially difficult problem, it’s important to implement this cleanly. We will define our own bit vector class to hold a large bit vector.

public static void checkDuplicates(int[] array) {
BitSet bs = new BitSet(32000);
for (int i = 0; i < array.length; i++) {
int num = array[i];
int num0 = num - 1; // bitset starts at 0, numbers start at 1
if (bs.get(num0)) {
System.out.println(num);
} else {
bs.set(num0);
}
}
}
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;
}
}

Comments:

bitset starts at 0 while numbers start at 1. I didn’t pay attention to that.

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