For those who missed it, the absence of "n" in these loops makes "do something" happen a constant number of times. So as written by the OP the loops are current

For those who missed it, the absence of "n" in these loops makes "do something" happen a constant number of times. So as written by the OP the loops are currently O(1) if for (int i = 0; i < 1000; i++) were instead for (int i = 0; i < n; i++) - You'd have O(n) If also for (int j = 1000; j >= 1; j = j / 2) were instead for (int j = n; j >= 1; j = j / 2) - Then you'd have O(n log n) OR for (int j = n; j >= 1; j -= 1) - O(n^2) OR for (int j = n * 1000; j >= 1; j -= 1) - this is STILL O(n^2)