DCFL or NOT

Question

Consider the follwoing Language

L₁= {0ⁿ1*0ⁿ | n>0} is DCFL or Not?

L₂= {0ⁿ1⁺0ⁿ | n>0} is DCFL or Not?

Comments

@Shubhanshu

it is actually (1,0/0)....i have wrote it by mistake.

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So, which means we are doing nothing with stack when 1 starts.

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yes..

Answer 1

L1 is a union of two languages L11 and L12, where 
L11 = $0^{2n}$ replace $1^*$ with epsilon
L12 = {$0^n1^+0^n$}  

L1 = L11 U L12 = Regular Language UNION DCFL = DCFL

L12 and L2 are same languages.

Hence, First and second both languages are DCFL.