The algorithm tells that, in the effort of solving $a^x \equiv b \text{ mod }N$:
Choose some $k \in \mathbb{N}$.
Create the baby list: $\{1,a,a^2,...,a^{k-1}\}$
Create the giant list: $\{ba^{-k},ba^{-2k},...,ba^{-rk}\}$ where $rk > N$.
Claim: If two lists have intersection, then this DLP has a solution.
$\textit{Proof:}$ Given that these two lists have an intersection, meaning that, for some $m,n$. \begin{align*} &a^n \equiv ba^{-mk} \text{ mod }N\\ & a^{mk+n} \equiv b \text{ mod }N\\ \end{align*} where $mk+n$ is $x$ as desired.
My question is, how do we know such solution is unique? or up to some equivalence? Is there any proof/counter-example for this?