majed - I'm not sure if this is more efficient than what you have, but it does produce the answer that you are looking for (assuming the the input chain is a string).
chain='455454545444544444445444444445444444544444445444444444444444434444343223';
setOf2 = chain(1);
startIndex = 1;
subchains = {};
for k=2:length(chain)
if ismember(chain(k),setOf2)
continue;
else
if length(setOf2) == 1
setOf2 = [setOf2 ; chain(k)];
else
subchains = [subchains chain(startIndex:k-1)];
startIndex = k;
setOf2 = chain(k);
end
end
end
subchains = [subchains chain(startIndex:end)];
subchains = [subchains chain(strtIdx:end)];
setOf2 is an array of the two distinct elements that are allowed in the subchain. We iterate over each member of the chain and check to see if it is a member of (contained within) the setOf2. If yes, then we continue to the next element. If not, then we have encountered a third unique element and so have completed the a subchain (which is added to the subchains cell array) and start a new one.
Try the above and see what happens!
