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.

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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.

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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);
		}
	}
}