带有条件内循环的 for 循环的 Big-O 表示法

Question

I am trying to understand the time complexity of this program, and why. I've made some notes of what I think it is, but I'm unsure if I understood it correct.

 public static int countSteps(int n) {
    int pow = 2;                             // O(1)
    int steps = 0;                           // O(1)
    for (int i = 0; i < n; i++) {            // O(n)
        if (i == pow) {                      // O(1)
            pow *= 2;                        // O(1)
            for (int j = 0; j < n; j++) {    // O(n)
                steps++;                     // O(1)
            }
        }
        else {
            steps++;                         // O(1)
        }
    }
    return steps; // O(1)
}

The inner loop spends a lot of time iterating through n every time the if-statement is triggered, does that affect the time complexity or is it still constant?

Answer 1

The outer loop means it's at least O(n). The inner loop will run only when i is a power of two, so it's O(n*log n).