| Note that over time, a pair of dots that are next to each other will always stay next to each other. | |
| If \(N\) is even and \(A_i = B_i\) for any \(i\), we know that an answer is not possible by definition. If \(N\) is odd, then there can be at most \(1\) place where \(A_i = B_i\) (the midpoint of the metalike-logo), so we can find it and get the answer in \(\mathcal{O}(N)\), or return \(-1\) if there's more than one \(i\) where \(A_i = B_i\). | |
| Otherwise, every \(A_i \ne B_i\), and we can first just think about each \(i\) as the lower vs higher dot. For a meta-like logo, the relationship always look like this: | |
| ``` | |
| (even N) (odd N) | |
| 3 2 3 5 6 4 -> LLLHHH 3 2 3 x 5 6 4 -> LLLxHHH | |
| 4 6 5 3 2 3 -> HHHLLL 4 6 5 x 3 2 3 -> HHHxLLL | |
| ``` | |
| Let's first consider even \(N\) by simulating the first pattern in reverse: | |
| ``` | |
| (t) (t-1) (t-2) (t-N/2) (t-N) | |
| LLLHHH LLLLHH LLLLLH LLLLLL HLLLLL HHLLLL HHHLLL | |
| HHHLLL HHHHLL HHHHHL HHHHHH LHHHHH LLHHHH LLLHHH | |
| (t-N-1) (t-2N) | |
| HHHHLL HHHHHL HHHHHH LHHHHH LHHHHH LLHHHH LLLHHH | |
| LLLLHH LLLLLH LLLLLL HLLLLL HLLLLL HHLLLL HHHLLL | |
| ``` | |
| Thus the pattern repeats after \(2N\) steps. If an answer exists, then at any given time: | |
| * for even \(N\), there's at most \(1\) place when \(A_i < B_i\) inverts to \(A_{i+1} > B_{i+1}\) (or the reverse). | |
| * for odd \(N\), the middle element eithers conform to its left or right side, so this still holds. | |
| Therefore, the midpoint of the metalike logo either be this inversion point, or the first/last element. Either way, there are \(\mathcal{O}(1)\) possible candidates for the midpoints. After we've found these candidates, we can simulate \(2N\) steps on the input arrays using a double-ended queue (updating the positions of the midpoint candidates at each step). If at any time a midpoint candidate actually shifts to the midpoint, we can do a check to see if \(A\) equals the reverse of \(B\). | |
| There are \(\mathcal{O}(1)\) midpoint candidates, and each can only reach the midpoint twice in our \(2N\) simulated time steps. Thus, the reverse condition is checked at most \(\mathcal{O}(1)\) times. The overall running time is \(\mathcal{O}(N)\). | |