@Absinthe My solution. This should be a somewhat space-optimized solution in ruby based off the modified concept of a Trie.

https://git.qoto.org/snippets/4

If i made this into a Radix Trie by compressing nodes with single children down I could reduce this further.

But since it does work I thought I'd share it as is. I'll update everyone if i decide to finish optimizing this particular solution.

It does however do a fairly decent job at compressing the tree by making sure no subtree is a duplicate of any other part of the tree (a node of any specific length/id will be the only node with that length.

For clarity I attached a picture from my notes of what the Trie would look like for the encoded string "12345" where the value inside each circle/node is the "length" value of that node, and the value attached to an arrow/vector/edge is the "chunk" associated with that link. The end result is any path from the minimum node (0) to the maximum node (5). This diagram does not include incomplete paths which my program does right now.

Incomplete paths can also be trimmed to further reduce the space. But since incomplete paths each add only a single leaf node to the tree, and might be useful for various use cases I decided to keep it.

#ruby #programming #toyprogrammingchallenge

@Absinthe

Linear in time and space by taking advantage of the Fibonacci numbers.

function count = interpretations(string)
ambig = string(1:end-1) == '1' | (string(1:end-1) == '2' & string(2:end) <= '6');
ambig &= string(1:end-1) ~= '0' & string(2:end) ~= '0' & [string(3:end) '1'] ~= '0';
ambig = [false ambig false];
consec = find(ambig(1:end-1) & !ambig(2:end)) - find(!ambig(1:end-1) & ambig(2:end));

fibo = zeros(max(consec), 1);
fibo(1:2) = [2 3];
for i = 3:max(consec)
fibo(i) = fibo(i - 1) + fibo(i - 2); end;
count = prod(arrayfun(@(x)fibo(x), consec)); end;