以Big-O表示法查找效率

Question

I was having problem with the following question

Consider the following nested loop construct. Categorize its efficiency in terms of the variable n using "big-o" notation. Suppose the statements represented by the ellipsis (...) require four main memory accesses (each requiring one microsecond) and two disk file accesses (each requiring one millisecond). Express in milliseconds the amount of time this construct would require to execute if n were 1000.

x = 1;
do
{
    y = n;
    while (y > 0)
    {
    ...
        y--;
    }
    x *= 2;
} while (x < n*n);

Answer 1

Inner loop with y is O(n).

Outer loop runs with x = 1, 2, 2^2, 2^3, ... 2^k < n * n. Hence it runs in O(log(n*n)) which is O(2 * log(n))

Hence complexity is O(n * log(n))

Answer 2

Just to add some more explanation to theother answer, a notable part of code is the x *= 2; ie a doubling. So this part is not linear. So you should be thinking log2.

Therefore, x will reach n*n in log2(n*n). = log2(n^2) = 2 x log2(n).

The y countdown is linear - so that is O(n)

There is a loop within a loop so you multiply both operations as in:

n * 2 x log2(n) = O(n * 2 * log2(n)). Then you take out constant factors to get: O(n * log2(n))