D. Miriany and Matchstick
constructive algorithms; dp; greedy
Miriany's matchstick is a \(2 \times n\) grid that needs to be filled with characters A or B. He has already filled in the first row of the grid and would like you to fill in the second row. You must do so in a way such that the number of adjacent pairs of cells with different characters\(^\dagger\) is equal to \(k\). If it is impossible, report so.\(^\dagger\) An adjacent pair of cells with different characters is a pair of cells \((r_1, c_1)\) and \((r_2, c_2)\) (\(1 \le r_1, r_2 \le 2\), \(1 \le c_1, c_2 \le n\)) such that \(|r_1 - r_2| + |c_1 - c_2| = 1\) and the characters in \((r_1, c_1)\) and \((r_2, c_2)\) are different.
The first line consists of an integer \(t\), the number of test cases (\(1 \leq t \leq 1000\)). The description of the test cases follows.The first line of each test case has two integers, \(n\) and \(k\) (\(1 \leq n \leq 2 \cdot 10^5, 0 \leq k \leq 3 \cdot n\)) – the number of columns of the matchstick, and the number of adjacent pairs of cells with different characters required.The following line contains string \(s\) of \(n\) characters (\(s_i\) is either A or B) – Miriany's top row of the matchstick.It is guaranteed that the sum of \(n\) over all test cases does not exceed \(2 \cdot 10^5\).
For each test case, if there is no way to fill the second row with the number of adjacent pairs of cells with different characters equals \(k\), output ""NO"". Otherwise, output ""YES"". Then, print \(n\) characters that a valid bottom row for Miriany's matchstick consists of. If there are several answers, output any of them.
In the first test case, it can be proved that there exists no possible way to fill in row \(2\) of the grid such that \(k = 1\). For the second test case, BABB is one possible answer.The grid below is the result of filling in BABB as the second row. \(\begin{array}{|c|c|} \hline A & A & A & A \cr \hline B & A & B & B \cr \hline \end{array}\) The pairs of different characters are shown below in red: \(\begin{array}{|c|c|} \hline \color{red}{A} & A & A & A \cr \hline \color{red}{B} & A & B & B \cr \hline \end{array}\)—————————————————\(\begin{array}{|c|c|} \hline A & A & \color{red}{A} & A \cr \hline B & A & \color{red}{B} & B \cr \hline \end{array}\)—————————————————\(\begin{array}{|c|c|} \hline A & A & A & \color{red}{A} \cr \hline B & A & B & \color{red}{B} \cr \hline \end{array}\)—————————————————\(\begin{array}{|c|c|} \hline A & A & A & A \cr \hline \color{red}{B} & \color{red}{A} & B & B \cr \hline \end{array}\)—————————————————\(\begin{array}{|c|c|} \hline A & A & A & A \cr \hline B & \color{red}{A} & \color{red}{B} & B \cr \hline \end{array}\)There are a total of \(5\) pairs, which satisfies \(k\).