Estimating $\theta$ based on censored data when $X_i\sim \text{Uniform}(0,\theta)$ with $\theta\ge 1$ Suppose $(X_i)_{1\le i\le n}$ are i.i.d $\text{Uniform}(0,\theta)$ random variables where $\theta \ge 1$. We observe $Y_i=\min(X_i,1)$ instead of $X_i$. I wish to estimate $\theta$ based on the data $(Y_i)_{1\le i\le n}$.
I think the likelihood takes the form
\begin{align}
L(\theta\mid \boldsymbol y)&=\prod_{i:x_i<1} \left(\frac1{\theta}\mathbf1_{0<y_i<\theta}\right)\prod_{i:x_i\ge 1}\left(1-\frac1{\theta}\right)
\\&=\theta^{-r}(1-\theta^{-1})^{n-r}\mathbf1_{0<y_{(n)}<\theta}\,,
\end{align}
where $r=\sum_{i=1}^n \mathbf1_{x_i<1}=\sum_{i=1}^n \mathbf1_{y_i<1}$ is the number of observations less than $1$.
If this is the correct likelihood, then a sufficient statistic for $\theta$ seems to be $T=(R,Y_{(n)})$. Or is the sufficient statistic simply $R$? Is the sufficient statistic complete? This would help to answer if there is UMVUE of $\theta$.
On the other hand, what is MLE of $\theta$? If I ignore the indicator $\mathbf1_{0<y_{(n)}<\theta}$ in the likelihood and assume $\theta\ne 1$, then differentiation leads to the stationary point $\hat\theta =\frac{n}{r}$. I am not sure if this is a valid answer.

It's redundant to have the $\textbf{1}_{0 < y_i < \theta}$ in the likelihood, because the product is already over samples where that is true. So you don't need that factor there. Then the likelihood is nicely differentiable and easy to optimize for $\theta$. And yes, your likelihood is correct without that factor.
When you do, you should get a result that seems intuitive. For example, if $\theta=2$, we should expect half the samples to be less than one. Think, in general, what proportion of samples (what $\frac{r}{n}$ in your notation) you should expect to see $<1$ for a given value of $\theta$, and see if that's consistent with your ML estimate $\hat\theta$.
