It is well known that 3-SAT is $\sf NP$-complete , but 2-SAT is in $\sf P$. Let there be a formula with $n-1$ clauses with 2 literals each and only 1 clause with 3 literals.
We can solve this case in polynomial time, separating and solving in a brute force manner the 3 literal clause and then for each satisfying assignment try to solve the rest $n-1$ 2-literal clauses. This method can work till $O(\log n)$ clauses with 3 literals. If we consider a more general case with e.g $\frac{n}{2}$ clauses with 2 literals and $\frac{n}{2}$ clauses with 3 literals does the problem remain $\sf NP$-complete?
It is a bit confusing because we have a subproblem approximately the same size, implying it is difficult and another one roughly the same size implying it is easy. Is there probably a better method than the one I proposed?