First I mark the walls between 2-3 and 3-4 clues as I mentioned before. Then I mark the same clues next to edge. Don't forget C5-C6 next to the wall.
As always I start the solving eith the largest numbers. In R10 the clue in 9. There are 12 cells there, but because of the walls at least C4 or C5 is empty. and the C9-C10 pair is same.
So R10C2 and R10C11 cannot be empty.
Similarly R2C7 and R11C7 are part of a rectangle. Then we can use the same-edge trick. Moreover R10C7 is marked, too.
Then I check the small numbers. 2s and 3s have the advantage, that all cells are in in group in that row.
We can mark many empty cells using these small numbers and the walls.
Such as R3C5 then R3C6 (same clues) and R3C9.
Then the top of C5 and C6.
R9C5 and R9C6 because of the 3 marked cells in the bottom of C7. The cloud in R1C7 continues to right.
R3C4 must be empty because of the same clues in R1-R2.
R12C4 is empty then R11C4, as well.
In C10, R3 is empty (C11 and C12 are same), then R1-R2. Moreover R9 is empty too, then R9C11-R9C12. Now the cloud in R10C11 continues to down 2 cells.
Now a trick in C10. The 3 cannot be in the top part. Because 7 in C11 cannot be 3+3+1.
And similarly R3C11 is empty. Now we can finish the whole bottom part.
Let's finish the puzzle. 6 in C8 cannot be 3+3 because of the 4 in C9. So R3C7-C8 are empty. And R4C7-C8-C9 are filled. This was the last relatively hard step.