The running time of an algorithm is $O(g(n))$ if and only if:

(A) its worst-case running time is $O(g(n))$ and its best-case running time is $\Omega(g(n))$.

(B) its worst-case running time is $\Omega(g(n))$ and its best-case running time is $O(g(n))$.

(C) $O(g(n))=\Omega(g(n))$.

(D) $o(g(n))\cap \omega(g(n))$ is non-empty.

Choose the correct answer from the options given below:

For an algorithm to have tight running time $\Theta(g(n))$, worst-case should be $O(g(n))$ and best-case should be $\Omega(g(n))$.

This condition is given in statement A.

Statement B reverses the required bounds.

Statement C is false because $O(g(n))$ and $\Omega(g(n))$ are not always equal.

Statement D is false because a function cannot be both smaller order and larger order than the same $g(n)$.