Given an input file with four billion nonnegative integers
FOLLOW UP: What if you have only 1 O MB of memory? Assume that all the values are distinct and we now have no more than one billion nonnegative integers.
There are a total of 232, or 4 billion, distinct integers possible and 231 nonnegative integers. Therefore, we know the input file (assuming it is ints rather than longs) contains some duplicates.
We have 1 GB of memory or 8 billion bits. Thus, with 8 billion bits, we can map all possible integers to a distinct bit with the available memory. The logic is as follows:
1. Create a bit vector (BV) with 4 billion bits. Recall that a bit vector is an array that compactly stores boolean values by using an array of ints (or another data type). Each int represents 32 boolean values.
2. Initialize BV with all Os.
3. Scan all numbers (num) from the file and call BV. set ( num, 1).
4. Now scan again BV from the 0th index.
5. Return the first index which has a value of 0.
The following code demonstrates our algorithm.
long numberOflnts = ((long) Integer.MAX_VALUE) + 1;
byte[] bitfield new byte [(int) (numberOfints / 8)];
String filename = void findOpenNumber() throws FileNotFoundException {
Scanner in = new Scanner(new FileReader(filename));
while (in.hasNextint()) {
int n = in.nextlnt ();
/* Finds the corresponding number in the bitfield by using the OR operator to
* set the nth bit of a byte (e.g., 10 would correspond to the 2nd bit of
* index 2 in the byte array). */
bitfield [n / 8] I= 1 « (n % 8);
}
for (int i= 0; i < bitfield.length; i++) {
for (int j = 0; j < 8; j++) {
/* Retrieves the individual bits of each byte. When 0 bit is found, print
* the corresponding value. */
if ((bitfield[i] & (1 << j)) == 0) {
System.out.println (i * 8 + j);
return;
}
}
}
}
Follow Up: What if we have only 10 MB memory?
It’s possible to find a missing integer with two passes of the data set. We can divide up the integers into blocks of some size (we’ll discuss how to decide on a size later). Let’s just assume that we divide up the integers into blocks of 1000. So, block O represents the numbers O through 999, block 1 represents numbers 1000 – 1999, and so on.
Since all the values are distinct, we know how many values we should find in each block. So, we search through the file and count how many values are between O and 999, how many are between 1000 and 1999, and so on. If we count only 999 values in a particular range, then we know that a missing int must be in that range.
In the second pass, we’ll actually look for which number in that range is missing. We use the bit vector approach from the first part of this problem. We can ignore any number outside of this specific range.
The question, now, is what is the appropriate block size? Let’s define some variables as follows:

Let rangeSize be the size of the ranges that each block in the first pass represents.

Let arrayS1ze represent the number of blocks in the first pass. Note that arraySize = 231/rangesize since there are 231 nonnegative integers.
We need to select a value for rangeSize such that the memory from the first pass (the array) and the second pass (the bit vector) fit.
First Pass: The Array
The array in the first pass can fit in 10 megabytes, or roughly 2 23 bytes, of memory. Since each element in the array is an int, and an int is 4 bytes, we can hold an array of at most about 2
21 elements.
So, we can deduce the following:
arraySize = (231/rangeSize)/<=231
rangeSize >= 231/221
rangeSize >= 210
Second Pass: The Bit Vector
We need to have enough space to store rangeSize bits. Since we can fit 223 bytes in memory, we can fit 26 bits in memory. Therefore, we can conclude the following:
211 <= rangeSize <= 226
These conditions give us a good amount of “wiggle room,” but the nearer to the middle that we pick, the less memory will be used at any given time.
The below code provides one implementation for this algorithm.
int findOpenNumber(String filename) throws FileNotFoundException {
int rangeSize = (1 << 20); // 2^20 bits (2^17 bytes)
/* Get count of number of values within each block. */
int[] blocks = getCountPerBlock(filename, rangeSize);
/* Find a block with a missing value. */
int blocklndex findBlockWithMissing(blocks, rangeSize);
if (blocklndex < 0) return 1;
/* Create bit vector for items within this range. */
byte[] bitVector = getBitVectorForRange(filename, blockindex, rangeSize);
/* Find a zero in the bit vector */
int offset findZero(bitVector);
if (offset < 0) return 1;
/* Compute missing value. */
return blockindex * rangeSize + offset;
}
/* Get count of items within each range. */
int[] getCountPerBlock(String filename, int rangeSize)
throws FileNotFoundException {
int arraySize = Integer.MAX_VALUE / rangeSize + 1;
int[] blocks = new int[arraySize];
Scanner in = new Scanner (new FileReader(filename));
while (in.hasNextint()) {
int value = in.nextint();
blocks[value / rangeSize]++;
}
in.close();
return blocks;
}
/* Find a block whose count is low. */
int findBlockWithMissing(int[] blocks, int rangeSize) {
for (int i= 0; i < blocks.length; i++) {
if (blocks[i] < rangeSize){
return i;
}
}
return 1;
}
/* Create a bit vector for the values within a specific range. */
byte[] getBitVectorForRange(String filename, int blockindex, int rangeSize)
throws FileNotFoundException {
int startRange = blockindex * rangeSize;
int endRange = startRange + rangeSize;
byte[] bitVector = new byte[rangeSize/Byte.SIZE];
Scanner in = new Scanner(new FileReader(filename));
while (in.hasNextint()) {
int value = in.nextint();
/* If the number is inside the block that's missing numbers, we record it */
if (startRange <= value && value < endRange) {
int offset = value  startRange;
int mask = (1 <<(offset% Byte.SIZE));
bitVector[offset / Byte.SIZE) I= mask;
}
}
in.close();
return bitVector;
}
/* Find bit index that is 0 within byte. */
int findZero(byte b) {
for (int i= 0; i < Byte.SIZE; i++) {
int mask= 1 << i;
if ((b & mask)== 0) {
return i;
}
}
return 1;
}
/* Find a zero within the bit vector and return the index. */
int findZero(byte[] bitVector) {
for (int i= 0; i < bitVector.length; i++) {
if (bitVector[i] != 0) {//If not all ls
int bitindex = findZero(bitVector[i]);
return i *Byte.SIZE+ bitindex;
}
}
return 1;
}
What if, as a followup question, you are asked to solve the problem with even less memory? In this case, we can do repeated passes using the approach from the first step. We’d first check to see how many integers are found within each sequence of a million elements. Then, in the second pass, we’d check how many integers are found in each sequence of a thousand elements. Finally, in the third pass, we’d apply the bit vector.