This a hand drawn solution to your problem, I'll explain the transitions,
Before you see the diagram, note that those are not 2 different DFAs, but I have drawn one more transition of states q1, q3, q5 to make the image more clear, as it is already very messed up.
q0 is the start state, machine has not seen anything until now.
On scanning the first symbol, suppose, it turns out to be 'a' d
d(q0, a) -> q1 machine (d is the transition function)
machine is in q1 now.
ifIf it sees another 'a' it will reach q2 which can beCAN BE the final state,
else it will reach q7 from where it will never reach any final state because 'a' has not repeated.
from q2, if machine see another 'a' it will go back to q1 else it will go to q3 (onon seeing a 'b' or q5 on seeing a 'c')
Same thing goes on for 'b' and 'c'
If afterAfter scanning the complete string, if the machine is in final state then the string is in the language else, it is not in the language.
This assumeassumes that the empty string epsilon is not in Double-Letter(L)
In that case q0 will be also be a final state.