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Write $L={\min }_i{X}_i$, $R={\max }_i{X}_i$ and $M={\max }_i|X_i|=\max \{-L,R\}$. The joint probability density factors asBy the Fisher-Neyman factorization theorem, $T=(L,R)$ is a sufficient statistic. The same factorization shows that the smaller statistic $M$ is also a sufficient statistic. This gives the sufficient statistic for a symmetric uniform sample.

In fact, the likelihood-ratio criterion for minimal sufficiency identifies $M$ as a minimal sufficient statistic: two sample points with the same $M$ have identical likelihoods for all $\theta$; with different $M$, their likelihoods have different parameter supports, and so are not proportional as functions of $\theta$. Equivalently, the condition that $f_{\theta }(x)=c{f}_{\theta }(y)$ for every $\theta$ and some positive constant $c$ holds exactly when their absolute maxima agree.

To see rigorously that $T$ cannot be recovered from $M$ almost surely, fix any $\theta >0$. The sample has a distribution invariant under simultaneous sign reversal. If $T=h(M)$ almost surely, the same identity would hold for the sign-reversed sample on another probability-one set. Since $M$ is unchanged by sign reversal, it would follow thatand hence $L=-R$ almost surely. But this event has probability zero for a finite sample from a continuous uniform distribution: it entails $X_i=-X_j$ for some pair of indices, including the possible case $X_i=0$. Therefore $T$ is not a function of the sufficient statistic $M$. The pair of extrema is not minimally sufficient, for every $n\ge 1$, even though it is sufficient.