Please explain this proof. Answer below.

Prove that the running time of an algorithm is theta (g(n)) if and only if its worst-case running time is O(g(n))and its best-case running time is omega (g(n)).

Now considering the worst-case running time as O(g(n)) and the best-case running time as 0(g(n)), it is enough to prove that the running time of an algorithm is (g(n)) o(g (n))-O(g (n))no(g(n)) f(n)-(g(n) f(n)-2((n)) From (1)& (2) it can be said f(n)-(g(n)) Hence proved that the running time of an algorithm is (g(n)) if and only if the worst-case running time is O(g(n)) and the best-case running time is (g(n))