No, the answer is A. "f(n) is O(g(n))" basically means the graph of f(n) will be bounded by the graph of g(n) at some point, and multiplied by some constant factor. And the notation is confusing because it's "=", but think of it as "<=" so O(n) = O(n^2) is true. so is O(n) = O(n^2) and O(n) = O(n^1000) or O(n) = O(2^n). Just like 1 < 2, and 1 < 4, 1 < 1000. Using the definition: O(nlogn) = O(n^2). We need to find c and n0 satisfying: f(n) <= cg(n) for n > n0. in our case nlogn <= cn^2 [for n > n0] take c = 1 and n0 = 10: nlogn <= 1n^2 Which is true, when n > 10, this above will always be true.