Algorithm Comparison with Theta-notation
If $T_1(n) = \Theta(n^2)$ and $T_2 = \Theta(n^3)$, then since $n^2 = o(n^3)$, it follows that $T_1(n) = o(T_2(n))$. In particular, $T_1(n) &lt; T_2(n)$ for large enough $n$. In fact, for every $C&gt;1$, $T_1(n) &lt; CT_2(n)$ for large enough $n$ (how large depends on $C$).