How is O(N) algorithm also an O(N^2) algorithm? | Solution

"Upper bound" means the algorithm takes no longer than (i.e. <=) that long (as the input size tends to infinity, with relevant constant factors considered). It does not mean it will ever actually take that long. Something that's O(n) is also O(n log n), O(n2), O(n3), O(2n) and also anything else that's asymptotically bigger than n. If you're comfortable with the relevant mathematics, you can also see this from the formal definition.