**Problem**

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?Solution

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.

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.

Then we print it out.

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); } }

Lets use a cleaner approach, by using BitSet in java. Note that bit set starts with 0, but number starts with 1.

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); } } }

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