For simplicity, we will consider the case \(N\) is odd. (When \(N\) is even, there are small differences, but the same main idea works.) Below, let \(M=\lfloor \frac{N}{2}\rfloor\).

An obvious upper bound for the answer is \(N-1\), only achieved by \(A=(1, 2, \ldots, N-1)\).

Considering the elements of this \(A\) from back to front, we can see that it has only two corresponding \(P\)’s:

\(P=(M+1,M,M+2,M-1,\ldots,1,N),(M+1,M+2,M,M+3,\ldots,N, 1)\).

If \(X=M+1\), we can print one of the above.

If \(X\neq M+1\), we have an upper bound of \(N-2\). Let us divide the elements other than \(X\) into two arrays \(L\) and \(R\), each of length \(M\), so that \(L_1<L_2<\ldots<L_M<R_1<R_2<\ldots<R_M\).

Now the upper bound is achieved by \(P=(X, L_M,R_1,L_{M-1},R_2,\ldots,L_1,R_M)\).