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

You know that log_2(5)=2.32. From this we note that 2^2 < 5 and 2^3 > 5.

Now look a matrix of possible answers:

j/i  0   1   2   3   4   5
 0   1   2   4   8  16  32
 1   5  10  20  40  80 160 
 2  25  50 100 200 400 800
 3 125 250 500 ...

Now, for this example, choose the numbers in order. There ordering would be:

j/i  0   1   2   3   4   5
 0   1   2   3   5   7  10
 1   4   6   8  11  14  18
 2   9  12  15  19  23  27
 3  16  20  24...

Note that every row starts 2 columns behind the row starting it. For instance, i=0 j=1 comes directly after i=2 j=0.

An algorithm we can derive from this pattern is therefore (assume j>i):

int i = 2;
int j = 5;
int k;
int m;

int space = (int)(log((float)j)/log((float)i));
for(k = 0; k < space*10; k++)
{
    for(m = 0; m < 10; m++)
    {
        int newi = k-space*m;
        if(newi < 0)
            break;
        else if(newi > 10)
            continue;
        int result = pow((float)i,newi) * pow((float)j,m);
        printf("%d^%d * %d^%d = %d\n", i, newi, j, m, result);
    }
}   

NOTE: The code here caps the values of the exponents of i and j to be less than 10. You could easily extend this algorithm to fit into any other arbitrary bounds.

NOTE: The running time for this algorithm is O(n) for the first n answers.

NOTE: The space complexity for this algorithm is O(1)