 Solution

Measuring the state from part (a) in the computational basis succeeds with probability $p_n={\sin }^2((2n+1)\theta )$. In the sparse-solution regime $0<M\le 2^{N-1}$, one has $0<\theta \le \pi /4$. The first success interval with $p_n\ge 1/2$ is $\pi /4\le (2n+1)\theta \le 3\pi /4$. Entering it requires $n\ge \pi /(8\theta )-1/2$, so the first half-probability Grover iterate isFor this integer, the angle is at least $\pi /4$ and less than $\pi /4+2\theta \le 3\pi /4$, so the rounding cannot skip the required interval. Every smaller nonnegative integer has angle below $\pi /4$ and hence success below $1/2$, proving minimality. For $M\ll 2^N$, the leading size is $n_{\min }\sim (\pi /8)\sqrt{2^N/M}$. If $M\ge 2^{N-1}$, the initial measurement already succeeds with probability at least one half and the smallest count is zero; if $M=0$ no search can succeed.