The question is rather simple, but I just can't find a good enough answer. I've taken a look at the most upvoted question regarding the Big-Oh notation, namely this: Plain English explanation of Big O It says there that: For example, sorting algorithms are typically compared based on comparison operations (comparing two nodes to determine their relative ordering). Now let's consider the simple bubble sort algorithm: for (int i = arr.length - 1; i > 0 ; i--) { for (int j = 0; j<i; j++) { if (arr[j] > arr[j+1]) { switchPlaces(...) } } } I know that worst case is O(n^2) and best case is O(n), but what is n exactly? If we attempt to sort an already sorted algorithm (best case), we would end up doing nothing, so why is it still O(n)? We are looping through 2 for-loops still, so if anything it should be O(n^2). n can't be the number of comparison operations, because we still compare all the elements, right? This confuses me, and I appreciate if someone could help me.