DCFL with prefix property have LR(0) grammar? There are two important theorems about LR(k) grammars and DCFL. Mentioned here.

A language has an LR(1) grammar iff it is DCFL.
A language has an LR(0) grammar iff it is DCFL and has prefix property.

I have counter example for 2nd theorem, plz see if it is valid counterexample.
(Ofcourse its not valid and I am wrong, But what wrong I am doing here ?)
DCFL without prefix property:    $\{b(ab)^n \mid n\geq0\}$
2nd theorem says it should not have $LR(0)$ grammar, but here it is-
$ S \rightarrow Sab \\S \rightarrow b$
PS: This is my first question here, sorry if I violated any protocol.
Your grammar is $\mathrm{LR}(0)$ by adding a $\mathrm{\$}$ symbol to the alphabet and a starting deduction rule $S'\rightarrow S\$$ when constructing the $\mathrm{LR}(0)$ automaton (and it is a standard construction).

Omitting syntactical purposes, every DCFL $\mathrm{L}$ can be parsed by an $\mathrm{LR}(0)$ grammar for $\mathrm{L}\$$. What makes $\mathrm{LR}(k)$ ($k&gt;0$) more practical is that it may include more syntactical information which is very important in compiler design. Otherwise, all these techniques would never be used in practice. In that case, $\mathrm{LL}(k)$ with local manipulation for expressions would be used instead, even with prohibitive cost of time and space.