Cracking the Coding Interview, 4 Edition - 150 Programming Interview Questions and Solutions

Solutions to Chapter 11 | System Design and Memory Limits
Follow Up: What if we have only 10 MB memory?
It’s possible to find a missing integer with just 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 0 represents the numbers
0 through 999, block 1 represents blocks 1000 - 1999, etc. Since the range of ints is finite, we
know that the number of blocks needed is finite.
In the first pass, we count how many ints are in each block. That is, if we see 552, we know
that that is in block 0, we increment counter[0]. If we see 1425, we know that that is in block
1, so we increment counter[1].
At the end of the first pass, we’ll be able to quickly spot a block that is missing a number. If
our block size is 1000, then any block which has fewer than 1000 numbers must be missing a
number. Pick any one of those blocks.
In the second pass, we’ll actually look for which number is missing. We can do this by creating
a simple bit vector of size 1000. We iterate through the file, and for each number that
should be in our block, we set the appropriate bit in the bit vector. By the end, we’ll know
which number (or numbers) is missing.
Now we just have to decide what the block size is.
A quick answer is 2^20 values per block. We will need an array with 2^12 block counters and
a bit vector in 2^17 bytes. Both of these can comfortably fit in 10*2^20 bytes.
What’s the smallest footprint? When the array of block counters occupies the same memory
as the bit vector. Let N = 2^32.
counters (bytes): blocks * 4
bit vector (bytes): (N / blocks) / 8
blocks * 4 = (N / blocks) / 8
blocks^2 = N / 32
blocks = sqrt(N/2)/4
It’s possible to find a missing integer with just under 65KB (or, more exactly, sqrt(2)*2^15
bytes).
1 int bitsize = 1048576; // 2^20 bits (2^17 bytes)
2 int blockNum = 4096; // 2^12
3 byte[] bitfield = new byte[bitsize/8];
4 int[] blocks = new int[blockNum];
5
6 void findOpenNumber() throws FileNotFoundException {
7 int starting = -1;
8 Scanner in = new Scanner (new FileReader (“input_file_q11_4.txt”));
2 0 3
Cracking the Coding Interview | Concepts and Algorithms