Question

Numbers whose only prime factors are 2, 3, or 5 are called ugly numbers.

Example:

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... 

1 can be considered as 2^0.

I am working on finding n^th ugly number. Note that these numbers are extremely sparsely distributed as n gets large.

I wrote a trivial program that computes if a given number is ugly or not. For n > 500 – it became super slow. I tried using memoization – observation: ugly_number * 2, ugly_number * 3, ugly_number * 5 are all ugly. Even with that it is slow. I tried using some properties of log – since that will reduce this problem from multiplication to addition – but, not much luck yet. Thought of sharing this with you all. Any interesting ideas?

Using a concept similar to Sieve of Eratosthenes (thanks Anon)

    for (int i(2), uglyCount(0); ; i++) {
        if (i % 2 == 0)
            continue;
        if (i % 3 == 0)
            continue;
        if (i % 5 == 0)
            continue;
        uglyCount++;
        if (uglyCount == n - 1)
            break;
    }

i is the n^th ugly number.

Even this is pretty slow. I am trying to find the 1500^th ugly number.

Answer 1

A simple fast solution in Java. Uses approach described by Anon..
Here TreeSet is just a container capable of returning smallest element in it. (No duplicates stored.)

    int n = 20;
    SortedSet<Long> next = new TreeSet<Long>();
    next.add((long) 1);

    long cur = 0;
    for (int i = 0; i < n; ++i) {
        cur = next.first();
        System.out.println("number " + (i + 1) + ":   " + cur);

        next.add(cur * 2);
        next.add(cur * 3);
        next.add(cur * 5);
        next.remove(cur);
    }

Since 1000th ugly number is 51200000, storing them in bool[] isn’t really an option.

edit
As a recreation from work (debugging stupid Hibernate), here’s completely linear solution. Thanks to marcog for idea!

    int n = 1000;

    int last2 = 0;
    int last3 = 0;
    int last5 = 0;

    long[] result = new long[n];
    result[0] = 1;
    for (int i = 1; i < n; ++i) {
        long prev = result[i - 1];

        while (result[last2] * 2 <= prev) {
            ++last2;
        }
        while (result[last3] * 3 <= prev) {
            ++last3;
        }
        while (result[last5] * 5 <= prev) {
            ++last5;
        }

        long candidate1 = result[last2] * 2;
        long candidate2 = result[last3] * 3;
        long candidate3 = result[last5] * 5;

        result[i] = Math.min(candidate1, Math.min(candidate2, candidate3));
    }

    System.out.println(result[n - 1]);

The idea is that to calculate a[i], we can use a[j]*2 for some j < i. But we also need to make sure that 1) a[j]*2 > a[i - 1] and 2) j is smallest possible.
Then, a[i] = min(a[j]*2, a[k]*3, a[t]*5).