Tricky Google interview question

Question

A friend of mine is interviewing for a job. One of the interview questions got me thinking, just wanted some feedback.

There are 2 non-negative integers: i and j. Gi

Answer 1

The algorithm implemented by user515430 by Edsger Dijkstra (http://www.cs.utexas.edu/users/EWD/ewd07xx/EWD792.PDF) is probably as fast as you can get. I call every number that is a form of 2^i * 5^j a "special number". Now vlads answer would be O(i*j) but with a double algorithm, one to generate the special numbers O(i*j) and one to sort them (according to the linked article also O(i*j).

But let's check Dijkstra's algorithm (see below). In this case n is the amount of special numbers we are generating, so equal to i*j. We are looping once, 1 -> n and in every loop we perform a constant action. So this algorithm is also O(i*j). And with a pretty blazing fast constant too.

My implementation in C++ with GMP (C++ wrapper), and dependancy on boost::lexical_cast, though that can be easily remove (I'm lazy, and who doesn't use Boost?). Compiled with g++ -O3 test.cpp -lgmpxx -o test. On Q6600 Ubuntu 10.10 time ./test 1000000 gives 1145ms.

#include 
#include 
#include 

int main(int argc, char *argv[]) {
    mpz_class m, x2, x5, *array, r;
    long n, i, i2, i5;

    if (argc < 2) return 1;

    n = boost::lexical_cast(argv[1]);

    array = new mpz_class[n];
    array[0] = 1;

    x2 = 2;
    x5 = 5;
    i2 = i5 = 0;

    for (i = 1; i != n; ++i) {
        m = std::min(x2, x5);

        array[i] = m;

        if (x2 == m) {
            ++i2;
            x2 = 2 * array[i2];
        }

        if (x5 == m) {
            ++i5;
            x5 = 5 * array[i5];
        }
    }

    delete [] array;
    std::cout << m << std::endl;

    return 0;
}