Explain why the statement, “The running time of algorithm $$A$$ is at least $$O(n^2)$$,” is meaningless.

Let us assume the running time of the algorithm is $$T(n)$$. Now, by definition, $$O$$-notation gives an upper bound for growth of functions but it doesn’t specify the order of growth.

Therefore, saying $$T(n) \ge O(n^2)$$ means growth of $$T(n)$$ is greater than some upper bound which is meaningless as we do not have any idea about what we are comparing it with.

For example, $$f(n) = 0$$ is $$O(n^2)$$ for all $$n$$. So, $$T(n) \ge f(n)$$ doesn’t tell us anything new as all running times are non-negative.