Is it possible to build DFA for odd-length words with 1 in the middle?
No, you can not. Proof by Pumping Lemma: Assume $L$ is regular and $n$ be the pumping length consider the word $w=0^n10^n$. Thus there are $x,y,z \in \{0,1\}^*$ with $|xy|&lt;n$ and $|y| &gt; 0$ such $xyz=w$. Then $\forall i\in \mathbb{N}_0: xy^iz \in L$. 

But due to its length constraint $x$ and $y$ consist of $0$ only and all are before the $1$. Thus $xy^2z = 0^{n+|y|}10^n \notin L$.

Thus, $L$ is not regular and can not be expressed by any DFA.