# Codeword length of a character with frequency more than $\frac{2}{5}$ in huffman coding

I read the following post. Now i encountered this question, what is codeword length of a character with frequency more than $$\frac{2}{5}$$ huffman coding? I think it can 2bit or 1bit, but i can't prove it. Any help for proving that appreciated.

Consider the classical algorithm that constructs the Huffman code and focus on the moment in which the singleton vertex $$v$$ corresponding to the character with frequency larger than $$\frac{2}{5}$$ was first merged with another tree $$T$$. At this point in time, either the frequency of $$T$$ was at least $$\frac{2}{5}$$ (which means that all trees had frequency at least $$\frac{2}{5}$$), or $$T$$ was the only tree with frequency smaller than $$\frac{2}{5}$$ (otherwise $$T$$ would have been merged with some other tree).
This means that there can be at most one additional tree in the forest other than $$v$$ and $$T$$.
The maximum depth of $$v$$ that can be obtained by combining (up to) $$3$$ trees is $$2$$.
Clearly both $$1$$ and $$2$$ are possible lengths of a codeword, as shown by symbol $$a$$ in the following two examples (frequencies are in blue).