DCFL or not??

Question

$L1 = \bigl\{a^mb^nc^pd^q   \mid m+q = n+p \bigr\}$
$L2 = \bigl\{a^mb^nc^pd^q   \mid m+p = n+q \bigr\}$

1. L1 is DCFL, L2 is not
2. L2 is DCFL, L1 is not
3. Both are not DCFL
4. Both are DCFL

Comments

2nd is also DCFL,

at every step it is exactly known(deterministically known) what operation to perform..

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They are right because at the end you find that in L1 stack is not empty to both are equal or not.

The key point of this question is 'state change in automata is in partcular sequence.'

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https://gateoverflow.in/15327/context-free-or-not

Answer 1

Both are DCFL.

L1:
Push $x$ on stack for $a$ and $d$, if the stack is empty or has $x$ on top. If however, the stack has $y$ on top, remove the $y$.
Push $y$ on stack for $b$ and $c$, if the stack is empty or has $y$ on top. If however, the stack has $x$ on top, remove the $x$.
The DFA accepts if the stack is empty at the end of the input.

 L2:
Push $x$ on stack for $a$ and $c$, if the stack is empty or has $x$ on top. If however, the stack has $y$ on top, remove the $y$.
Push $y$ on stack for $b$ and $d$, if the stack is empty or has $y$ on top. If however, the stack has $x$ on top, remove the $x$.
The DFA accepts if the stack is empty at the end of the input.

Comments

State change in automata in sequence.First 'a' push then 'b' pop then 'c' pop and finally push 'd'  then how it is possible to check at the end stack is empty?

I think L1 is not DCNF and L2 is DCNF  i.e. option 2

Answer 2

option 2
Because in L1 automata, you find at the end of operation stack actually not empty to compare it.