We focus on the last character of the string.

[1] When the last character of \(S\) is A

In this case, it can be shown that the last character of \(f(S)\) is A. First, it is obvious when \(S={}\)AA...AA. When \(S\) contains B, let \(x\) be the maximum \(i\) such that \(S_i={}\)B. As \(S\) ends with A, we have \(x\neq N\), and the last character of \(f(S)\) will be \(S_{x+1}={}\)A. Thus, the claim has been shown.

From this, if we replace a string \(S\) ending with A with \(f(S)\) \(N-1\) times, it is seen that \(S\) becomes A.

[2] When the last character of \(S\) is B

When \(S={}\)BB...B, the answer is clearly B. Let’s consider \(S\) other than this.

When will the last character of \(f(S)\) be A for the string \(S\) ending with B? For that, it is necessary that there is an \(i\) such that \(S_i={}\)B and \(S_{i+1}={}\)A. Let’s consider two cases according to whether such an \(i\) exists.

[2-1] If no such \(i\) exists

No such \(i\) exists only when \(S\) is a concatenation of \(x\ (1\leq x)\) As and \(y\ (1\leq y)\) Bs. In this case, \(f(S)\) will be a concatenation of \(x-1\) As and \(y\) Bs. Specifically, \(f(S)\) also satisfies the following conditions:

• the last character is B;
• there is no \(i\) such that the \(i\)-th character is B and the \(i+1\)-th character is A.

Therefore, even after replacing \(S\) with \(f(S)\) \(N-1\) times, the last character is B, and \(S\) becomes B.

[2-2] If such \(i\) exists

Let \(x\) be the maximum such \(i\). \(S\) is structured as \(S={}\)...BA...ABB...BB. For \(f(S)\) for such \(S\), let \(y\) be the number of consecutive Bs at the end of \(S\), and the following holds:

• there are exactly \(y-1\) consecutive Bs at the end of \(f(S)\);
• the \(y\)-th character from the end of \(f(S)\) is \(S_{x+1}={}\)A;
• \(S_{i+1}={}\)B for the maximum \(i\) such that \(S_{i}={}\)A, and this B will be in \(f(S)\) as the \(y+1\)-th character or beyond from the end.

From the second and third points, it can be seen that also in \(f(S)\), there is an \(i\) such that the \(i\)-th character is B and the \(i+1\)-th character is A. Furthermore, the number of consecutive Bs at the end of \(f(S)\) is one less than that of \(S\). Therefore, if you replace \(S\) with \(f(S)\) \(y\) times, the last character of \(S\) becomes A.

From this, it can be seen that if we replace such a string \(S\) with \(f(S)\) \(N-1\) times, \(S\) becomes A.

(Modified the first draft written by GPT-4.)