My signal is x(n) = {3,2,1,-1} for n=0,1,2,3
Using the definition of the DFT I found that $$ X(k) =\sum_{n=0}^{3} x(n)e^{-j\frac{2πnk}{N}} = 3 + 2e^{-j\frac{2πk}{N}} + e^{-j\frac{4πk}{N}} - e^{-j\frac{6πk}{N}}$$
Which should be correct right? Also if I'm not wrong we need 4 complex multiplications to come to this result.
Now my question is how can I get the DFT of x(n) if I have g(n) = {1,-3,-2,-1} and G(k) = {-5, 3+2j, 3, 3-2j}, while using the properties of DFT?
Is it possible to say that g(n) = -x(n+1) ?
Considering periodic continuation outside the interval $[0,N-1]$, you can state that
$$x[n]=-g[n+1]\tag{1}$$
from which it follows that
$$X[k]=-G[k]e^{j\frac{2\pi}{N}k}\tag{2}$$
With $N=4$ you get from $(2)$
$$X[k]=-G[k]j^k\tag{3}$$
which can be easily verified given the concrete example in your question.