For the loop for (i = 0; i < n; i += k), return how many times its body runs. Finding a loop's complexity starts with counting its iterationsIterationOne pass through a loop body.. Here i starts at 0 and jumps by k, so the body runs about n / k times — exactly ⌈n / k⌉. When k is a fixed constant this is O(n); the step size changes the count by a constant factor, not the growth rate.