Recurrence Question:T(n) = T(n − √ n) + T( √ n) + θ(n)
The solution is $T(n) = \Theta(n^{3/2})$.

For the lower bound, note that $T(n) \geq T(n-\sqrt{n}) + \Theta(n)$. Iterating this for $\sqrt{n}/2$ steps, the value of the parameter never drops below $n/2$, for a total contribution of $\Omega(n^{3/2})$.

For the upper bound, let us notice first that
$$(n-\sqrt{n})^{3/2} \approx n^{3/2} - \sqrt{n} \frac{d}{dn} n^{3/2} = n^{3/2} - \Theta(n). $$
This allows us to prove the upper bound $T(n) = O(n^{3/2})$ by induction.