Find a minimal sufficient statistic for $U(\theta,\theta+c)$, where $(\theta,c)$ unknown.

(Nobody answers so I post this answer which I am not sure is correct)

We show $(Y_{(1)},Y_{(n)})$ is a sufficient complete statistic, which implies $(Y_{(1)},Y_{(n)})$ is minimal sufficient. We first give the p.d.f of $(Y_{(1)},Y_{(n)})$ $$f(y_1, y_n,\theta,c)=\frac{n(n-1)}{c^n}(y_n-y_1)^{n-1},\quad\forall \theta\leq y_1\leq y_n\leq \theta+c\text{ and } (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$$

Then, for any function $g(x_1,x_n)$ so that $\mathbb{E}_{\theta,c}\left[g(y_1,y_n)\right]=0,\forall (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$, we have $$0=\frac{n(n-1)}{c^n}\int_{\theta}^{\theta+c}\int_{\theta}^{y_n}g(y_1,y_n)(y_n-y_1)^{n-2} dy_1dy_n,\forall (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$$ The integral area of $(y_1, y_n)$ is a triangle with vertices $(\theta,\theta)$, $(\theta,\theta+c)$ and $(\theta+c,\theta+c)$. With varying $\theta\in\mathbb{R}$ $c\in\mathbb{R}^+$, these triangles generate the Beral $\sigma$-algebra of $\mathcal{B}=\{(x,z)\in\mathbb{R}^2:x\leq z\}$. Thus $$0=\frac{n(n-1)}{c^n}\int_{A}g(y_1,y_n)(y_n-y_1)^{n-2} d(y_1,y_n),\text{ for any Borel set }A\subset\mathcal{B}.$$ This means $$g(y_1,y_n)(y_n-y_1)^{n-2}\equiv0,a.e.\iff g\equiv 0,a.e.$$ Thus we conclude $(Y_{(1)},Y_{(n)})$ is a complete statistic for $(\theta,c).$

Since $(Y_{(1)},Y_{(n)})$ is sufficient by factorization theorem, we conclude $(Y_{(1)},Y_{(n)})$ is minimal sufficient.