We save the input as a matrix ~mat~. Let the starting island have coordinates ~(ox, oy)~, so ~mat[ox][oy] =~ o. We will try to send the ducks to each of the four directions, yielding four paths. For each of these paths, we count through how many cells it passes before reaching x, or say the distance is infinite if the path ends in . or o.

Implementation: we run the following algorithm four times, starting from ~(ox+1, oy)~, ~(ox-1, oy)~, ~(ox, oy+1)~ and ~(ox, oy-1)~. We maintain the current duck position as ~(px, py)~, and the length of the path in a variable ~d~ initialized to ~0~. If ~mat[px][py]~ is in <>^v, increment ~d~ and move ~(px, py)~ in the corresponding direction. (We can use dict or std::map, mapping the ordered pairs which describe the direction to characters, e.g. {'>': (0, 1)}). If ~mat[px][py]~ equals o or ., set ~d = \infty~ and stop processing this path. We are done if ~mat[px][py]~ equals x.

After the previous part, we have the distances ~d_N~, ~d_E~, ~d_W~ and ~d_S~. If all ~d_i = \infty~, there is no solution and we print :(.

Else, we print :) and the letter ~i~ for which ~d_i~ is minimal, breaking ties alphabetically.