Ans should be 2
L1 : CFL
Reason: Context free grammar for this language
S--> 0S1 / 1S0 / 0S0 / 1S1 / 0
(ODD) 0 (ODD) = ODD length strings
(Even) 0 (Even) = ODD length strings
L2: CFL ( easily you can visualise )
L3: CSL ( because here Total comparison >= 2 for same symbol. 1comparison: first time number 'a' should be equal to last time number of 'a'. 2 comparison: 'b' in middle should not equal to the number of 'a'.)
Please let know if I am wrong.
okk, Let's take some assumption to understand this language.
i!=j (assume A) "OR" j!=k ( assume B)
we know that A OR B must be true if at least one is true either( A or B or BOTH). Here PDA is possible but DPDA is not. For making both simultaneously true in the PDA, you have the choice to accept the string either A side or B side in PDA.