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 \(\Omega(g(n))\).

Let’s assume that the running time of the algorithm is \(T(n)\). If \(T(n) = \Theta(g(n))\), then for \(n \ge n_0\),

\[0 \le c_1 g(n) \le T(n) \le c_2 g(n)\]As \(0 \le T(n) \le c_2 g(n)\) for \(n \ge n_0\), \(T(n) = O(g(n))\), i.e. \(T(n)\) is upper bounded by \(O(n)\). In other words, worst-case running time of the algorithm is \(O(n)\).

And as \(0 \le c_1 g(n) \le T(n)\) for \(n \ge n_0\), \(T(n) = \Omega(g(n))\), i.e. \(T(n)\) is lower bounded by \(\Omega(n)\). In other words, best-case running time of the algorithm is \(\Omega(n)\).

Similarly we can prove the reverse as we did in Exercise 3.1-5.