Assuming we use decoder/demux since we need the ENABLE to combine 2 decoders.
Let’s number bits from $(LSB \rightarrow MSB)$ as:
$\Rightarrow n-\text{bits} = \{\underline{b_0, b_1, …, b_k}, \underline{b_{k+1}, … b_n}\}$
Let’s use $decoder_1 = k \times 2^k$ for the $k$ least significant bits
i,e., $decoder_1$ is connected to bits $\{b_0, b_1, …, b_k\}$
Since $decoder_1$ will always be active,
$\text{ENABLE}=1$
Let’s use $decoder_2 = (n-k) \times 2^{n-k}$ for the $(n-k)$ most significant bits
i,e., $decoder_2$ is connected to bits $\{b_{k+1}, b_{k+2}, …, b_n\}$
Since $decoder_2$ will only be active when there is atleast one active MSB
$\text{ENABLE} = (b_{k+1} + b_{k+2} + … b_{n})$
At any given time, either one or both the decoders will be active.
Best Case:
There is no MSB active, $\therefore$ only $decoder_1$ is active.
$T_n = \Omega(2^k)$
Worst Case:
There is atleast one active MSB, $\therefore$ both $decoder_1, decoder_2$ is active
$T_n = O(max\{2^k , 2^{n-k}\})$
Average Case:
In average case, both the decoders will be active.
$T_n = \Theta(max\{2^k , 2^{n-k}\})$
To get minimum address decoding time, i.e., minimum worst/average time complexity,
we need to find the minimum value of $k , 0 \leq k \leq n$ such that
$max\{2^k , 2^{n-k}\}$ is minimum
which is possible when both are equal,
i.e., $2^k = 2^{n-k}$
$\Rightarrow k = n-k$
$\Rightarrow k = \frac{n}{2}$