#toyprogrammingchallenge
Another Freebie...
This problem was asked by Facebook.
Given the mapping a = 1, b = 2, ... z = 26, and an encoded message, count the number of ways it can be decoded.
For example, the message '111' would give 3, since it could be decoded as 'aaa', 'ka', and 'ak'.
You can assume that the messages are decodable. For example, '001' is not allowed.
🇳🇱
@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.
@Absinthe BTW i intentionally chose to list them not count them. I could probably optimize further if I counted only. In fact that would be trivial, so i think ill solve that too.
lol not efficient if **all** you need is the count, but it does work of course.
@billstclair @Absinthe Time-complexity has nothing to do with raw-speed. If you do the brute force solution and actually list all solutions with a copy approach (the absolute simplest least efficient way) then your dealing with exponential space complexity. Good luck computing large datasets on even the beefiest of machines where a linear solution would attack it with ease.
@billstclair @freemo how does that solve it? Am I missing something?
My Elm solution. Change the input sequence of numbers from 1 to 9, and the count plus all the possible decodings will be output instantly.
https://ellie-app.com/6Q5BTjx8CGta1
#toyprogrammingchallenge
@billstclair
I don't know about that word "instantly" there but cool solution ;)
@Absinthe
@billstclair @Absinthe @freemo
This solution gives wrong answers in certain cases. E.g. "10" should return one solution, the letter "j', but instead it gives zero. I'm suspicious of the inequalities on line 95 (strictly greater than zero AND strictly less than twenty-six permits the integers 1-25, which looks like a fencepost error given a 26-character alphabet).
Good catch, elm isnt one of the languages i mess with so i didnt run it myself and only got a vague idea of what is going on.
Octave solution
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;
@Absinthe #toyprogrammingchallenge I did this with recursion (plus caching). Will create a repo tomorrow, but the lru_cache in Python is pretty awesome. The recursive solution starts to slow down significantly as the input strong gets longer, but caching the intermediate results keeps it super fast.
@Absinthe Since the main complexity issue here becomes space complexity im going to tackle this as a space optimized solution. Sould also be pretty close to ideal on time complexity if i do it right too.