这个循环是O（nlog（n））吗？

Question

I have a nested for loop that I am trying to analyze the efficiency of. The loop looks like this:

int n = 1000;
for (int i = 0; i < n; i++) {
    for (int j = 0; j < i; j++) {
        System.out.print("*");
    }
}

I don't believe that this algorithm is O(n^2) because the inner loop does not run n times, it only runs i times. However, it certainly is not O(n). So I hypothesize that it must be between the two efficiencies, which gives O(nlog(n)). Is this accurate or is it really a O(n^2) algorithm and I'm misunderstanding the effect the inner loop has on the efficiency?

Answer 1

Your algorithm will run a triangular number of times:

n * (n + 1) / 2

In the above case, n = 999 because the first j loop doesn't run:

(999 * 1000) / 2 = 499500

It is lower than n**2 , but it still is O(n**2) , because n * (n + 1) / 2 is n**2 / 2 + n / 2 . When n is large, you can ignore n / 2 compared to n**2 / 2 , and you can also ignore the constant 1 / 2 factor.

Answer 2

I kind of get your doubts, but try to think in this way: what value will i have in the worst case scenario? Answer is n-1 , right? So, as the complexity is evaluated by considering the worst case scenario it turns out that it is O(n^2) as n * (n-1) ~ n^2 .

Answer 3

The number of iterations is sum from i=0 to n-1 (sum from j=0 to i-1 (1)) . The inner sum is obviously equal to i . sum from i=0 to n-1 (i) = n * (n-1) / 2 = O(n^2) is well known.