In this case D & E have the lowest frequencies and so are allocated 0 and 1 respectively and grouped together with a combined probability of 0.282. The lowest pair now are B and C so they're allocated 0 and 1 and grouped together with a combined probability of 0.333. This leaves BC and DE now with the lowest probabilities so 0 and 1 are prepended to their codes and they are combined. This then leaves just A and BCDE, which have 0 and 1 prepended respectively and are then combined. This leaves us with a single node and our algorithm is complete.