Editorial for SAC '22 Code Challenge 5 P2 - Querying Extensions

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Author: maxcruickshanks

Subtask 1

It suffices to hard code the output for every $N$ ~N~ and left or right.

Time Complexity: $\mathcal{O}(1)$ ~\mathcal{O}(1)~

Subtask 2

Realize that this is a parity problem:

If $N \equiv 1 \pmod 2$ ~N \equiv 1 \pmod 2~, then the output is the same as the original swaying direction.
Otherwise, if $N \equiv 0 \pmod 2$ ~N \equiv 0 \pmod 2~, then the output is the opposite of the original swaying direction.

Output the direction based on the two cases.

Time Complexity: $\mathcal{O}(1)$ ~\mathcal{O}(1)~

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