The Efficient Steps (Editorial)

Tutorial

To minimize the total number of moves, we should subtract the largest possible value, which is $3$, as often as possible.

Let's divide $N$ by $3$ and analyze the remainder to find the optimal strategy. There are exactly 3 possible cases:

1. N \pmod 3 = 0: We can perfectly reduce $N$ using only $3$s. We subtract $3$ exactly $N/3$ times. Total moves = $N/3$.

2. N \pmod 3 = 2: We subtract $3$ as many times as possible ($\lfloor N/3\rfloor$ times) and are left with exactly $2$. We then subtract $2$ one time. Total moves = \lfloor N/3 \rfloor + 1.

3. N \pmod 3 = 1: If we greedily take out $3$s, we are left with $1$, which we cannot clear because we can only subtract $2$ or $3$. Instead, we subtract $3$ one less time than the maximum, leaving a remainder of 3 + 1 = 4. We can then perfectly clear this $4$ by subtracting $2$ two times. Total moves = (\lfloor N/3 \rfloor - 1) + 2 = \lfloor N/3 \rfloor + 1.

Notice that for any $N$ that is not a perfectly divisible by $3$, the minimum number of moves is always exactly \lfloor N/3 \rfloor + 1. We can compute and print this for each testcase.

Complexity Time Complexity: $\mathcal{O}(1)$ per testcase to compute the answer using basic arithmetic, resulting in $\mathcal{O}(T)$ overall. Space Complexity is $\mathcal{O}(1)$.