Exactly, the better greedy aprroach is using DP, then $$$dp_i$$$ means the answer when we consider first $$$i$$$ elements, it can be updated in two ways, the last block which has $$$ith$$$ position is divisible by 3, or not, if not then $$$dp_i$$$ will be equal to $$$dp_{i-1}$$$, and otherwise cut the smallest block which is divisible by 3, then $$$dp_i$$$ will be $$$dp_j+1$$$(finding the smallest block can be done whit a precalculation with $$$O(n)$$$ time).

The whole problem will be $$$O(n)$$$.

Thanks a lot poduingi