 Solution

The algorithm returns a uniformly random marked input with certainty. The Born rule applied to the final state gives, for each individual basis label $y$,Summing over the $S_1=N/4$ marked inputs gives $\mathrm{Pr}(f(Y)=1)=1$. The outcome is a marked input label, not merely the bit indicating whether some marked input exists. There is one use of the supplied phase oracle, together with the specified preparation, reflection and measurement.

The exact success relies on the marked fraction $r=S_1/N=1/4$. More generally, the mean amplitude immediately after the oracle is $(1-2r)/\sqrt{N}$. Inversion about the mean sends an unmarked amplitude to $(1-4r)/\sqrt{N}$ and a marked amplitude to $(3-4r)/\sqrt{N}$. Thus a single iteration has marked success probability $r(3-4r{)}^2$. At $r=1/4$ the unmarked amplitudes vanish and the marked amplitudes become $2/\sqrt{N}$, giving the one-step exact Grover search with a quarter of the inputs marked. No claim of one-query certainty for an arbitrary marked fraction is needed.