“Unique” Random Numbers technical interview question

Dan Karp used to ask this question at Zimbra. I’ve always loved it.

1. warmup

Given this API from the java.lang standard Math class in the JDK:

* Returns a double value with a positive sign,
* greater than or equal to 0.0 and less than 1.0.
* Returned values are chosen randomly with
* uniform distribution from that range.
public static double Math.random();

Your task is to implement a random integer in the range [0,999]. Don’t overthink it; this is just a warmup.


public class Foo {
public static int randomInt1000() {
return (int) (Math.random() * 1000);

Idea is to truncate a number like 123.99748 to 123 by casting a double to an int. No need to use Math.floor(), that just wastes a function call and does some floating-point arithmetic.

Some candidates will want to avoid hard-coding the 1000 constant, and will create a function signature like this:

public static int randomInt(int N);

Great. Either answer is totally fine.

2. introduce the idea of unique-ness

Things are about to get harder.

If I call your Foo.randomInt1000() function a bunch of times, I’ll get values like this:

316, 675, 22, 671, 724, 394, 31, 177, 997, …, 82, 671, 810, 433, …

If you call the function enough times, of course you’ll see the same number more than once.

Create a new function called uniqueRandomInt1000() that returns a random integer in the range [0,999] inclusive but doesn’t return duplicates. Your function will need to keep some state, and it’s up to you how to manage that state.

We’re looking for (1) a correct solution to the problem, and (2) an efficient solution. Correctness is more important than efficiency. But efficiency matters, too.

Edge case: if the caller calls uniqueRandomInt1000() more than 1000 times, the implementation needs to signal in some way that there are no more unique random numbers that can be returned because every number in the range [0,999] has been returned to the caller. Either return -1 or throw an Exception.

3.0. a naive, correct, but O(unbounded) solution

Keep an array of booleans or a java.util.BitSet, or a Set<Integer> of every number you’ve returned as an instance variable for your class. Every time the method gets called, generate a random integer in the range 0,999 and then check your instance variable to see if that number has been “seen” by the caller. If so, randomly generate another integer in the range 0,999 and repeat until you find an “unseen” number. Mark that number as seen and return to the caller.

3.1. a correct, O(N^2) solution using a List

Key idea here to avoid looping is to generate a random index in an ever-smaller range of numbers, and use that index to select a number from a LinkedList or an ArrayList.

In the ctor you populate a List with 1000 elements – every number between 0,999. That’s O(N) time.

Then, in the function, you generate a random index in the range 0,size()-1. size() is initially 1000, but shrinks over time as you remove() elements from the List.

get() followed by remove() is always O(N). With an ArrayList, you get constant time indexing into the position you want to retrieve (get()), but then you spend linear time copying elements over to compact the list (remove()). With a LinkedList you get linear scanning to move to the node at a particular index (get()), and then you get constant time to tidy up the list to remove that element (remove()).

If you ask for N random numbers from the class you get N * O(N) == O(N^2) cost.

3.2. solution using Java collections O(N)

This solution does O(2N) in the ctor and then O(1) for each call. It’s kinda “cheating” a little bit because all of the hard algorithmic work is done by a single function call Collections.shuffle().

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class UniqueRandom {
private List numbers;
private int pos = 0;

public UniqueRandom(int n) {
numbers = new ArrayList(n);
for (int i = 0; i < n; i++) { numbers.add(i); } Collections.shuffle(numbers); } public int nextInt() { if (pos == numbers.size()) { return -1; } return numbers.get(pos++); } public static void main(String[] args) { UniqueRandom ur = new UniqueRandom(35); for (int i = 0; i < 40; i++) { System.out.println(ur.nextInt()); } } }[/code]

3.3. solution using an int[]

This solution does O(1N) in the ctor and then O(1) for each call.

public class UniqueRandom2 {
private int[] numbers;
private int avail;

public UniqueRandom2(int n) {
avail = n;
numbers = new int[avail];
for (int i = 0; i < avail; i++) { numbers[i] = i; } } public int nextInt() { if (avail == 0) { return -1; } int idx = (int) (Math.random() * avail); int retval = numbers[idx]; numbers[idx] = numbers[--avail]; return retval; } public static void main(String[] args) { UniqueRandom2 ur = new UniqueRandom2(1000); for (int i = 0; i < 40; i++) { System.out.println(ur.nextInt()); } } }[/code] Thanks to Ryan Hamilton for reminding me about this so I had an excuse to dig up my sample code and post it online.