sure, in reception of QPSK, you receive bit pairs as they are transmitted, and detangle them in the reverse manner that they were grouped together.
what's really cool about OQPSK is that the order of the bits going in can naturally determine the order of bit changing in OQPSK. in fact, it can be made into a DSP modulation system without the need for grouping bits together in pairs.
define the bitstream $a[n] \in \{0,1\}$, and the bipolar binary signal as:
$$ x[n] \ = \ (-1)^{a[n]} \ = \ 1 - 2 a[n]\ \in \ \{+1,-1\} $$
then define these two gating functions:
even samples:
$$ g[n] \ = \ \tfrac{1}{2} \left(1 + (-1)^n \right) \ = \ \tfrac{1}{2} \left(1 + e^{j \pi n} \right) \ = \ \begin{cases}
1, & n\text{ even} \\
0, & n\text{ odd}
\end{cases} $$
odd samples:
$$ 1 - g[n] \ = \ g[n-1] \ = \ \tfrac{1}{2} \left(1 - (-1)^n \right) \ = \ \tfrac{1}{2} \left(1 - e^{j \pi n} \right) \ = \ \begin{cases}
0, & n\text{ even} \\
1, & n\text{ odd}
\end{cases} $$
the I/Q quadrature pair is:
$$ i[n] \ = \ g[n] \ x[n] \ + \ (1-g[n]) \ x[n-1] $$
$$ q[n] \ = \ (1-g[n]) \ x[n] \ + \ g[n] \ x[n-1] $$
note that $i[n+1]=i[n]=x[n]$ for even $n$ and $q[n+1]=q[n]=x[n]$ for odd $n$. so for either $i[n]$ or $q[n]$, the bit rate is half the bit rate is for $a[n]$ or $x[n]$. so the bandwidth needed for the analog reconstructed signals $i(t)$ and $q(t)$ need only be half of the bandwidth needed for the reconstruction of $x(t)$ from $x[n]$ is.
then the discrete-time OQPSK modulating signal is
$$ s[n] \ = \ i[n] \ + \ j \ q[n] $$
returning to the real $i[n]$ and $q[n]$:
$$\begin{align}
i[n] \ & = \ g[n] \ x[n] \ + \ (1-g[n]) \ x[n-1] \\
& = \ \tfrac{1}{2} \left(1 + e^{j \pi n} \right) \ x[n] \ + \ \tfrac{1}{2} \left(1 - e^{j \pi n} \right) \ x[n-1] \\
& = \ \tfrac{1}{2} (x[n] + x[n-1]) \ + \ \tfrac{1}{2} e^{j \pi n} (x[n] - x[n-1]) \\
\end{align}$$
$$\begin{align}
q[n] \ & = \ (1-g[n]) \ x[n] \ + \ g[n] \ x[n-1] \\
& = \ \tfrac{1}{2} \left(1 - e^{j \pi n} \right) \ x[n] \ + \ \tfrac{1}{2} \left(1 + e^{j \pi n} \right) \ x[n-1] \\
& = \ \tfrac{1}{2} (x[n] + x[n-1]) \ - \ \tfrac{1}{2} e^{j \pi n} (x[n] - x[n-1]) \\
\end{align}$$
the OQPSK modulating signal is
$$\begin{align}
s[n] \ & = \ i[n] \ + \ j \ q[n] \\
& = \ \tfrac{1}{2} (x[n] + x[n-1]) \ + \ \tfrac{1}{2} e^{j \pi n} (x[n] - x[n-1]) \ + \ j \ \left( \tfrac{1}{2} (x[n] + x[n-1]) \ - \ \tfrac{1}{2} e^{j \pi n} (x[n] - x[n-1]) \right) \\
& = \ \frac{1+j}{2} (x[n] + x[n-1]) \ + \ \frac{1-j}{2} e^{j \pi n} (x[n] - x[n-1]) \\
\end{align}$$
now compute the Discrete-time Fourier Transform (DTFT)
$$\begin{align}
S(\omega) \ & = \ \frac{1+j}{2} \left(X(\omega) + e^{-j\omega}X(\omega) \right) \ + \ \frac{1-j}{2} \left(X(\omega-\pi) - e^{-j(\omega-\pi)}X(\omega-\pi) \right) \\
& = \ \frac{1+j}{2} X(\omega) \left(1 + e^{-j\omega} \right) \ + \ \frac{1-j}{2} X(\omega-\pi) \left(1 - e^{-j(\omega-\pi)} \right) \\
& = \ \frac{1+j}{2} X(\omega) \left(1 + e^{-j\omega} \right) \ + \ \frac{1-j}{2} X(\omega-\pi) \left(1 + e^{-j\omega} \right) \\
& = \ \frac{1 + e^{-j\omega}}{2} \Big( (1+j)X(\omega) \ + \ (1-j)X(\omega-\pi) \Big) \\
& = \ e^{-j\omega/2}\frac{e^{j\omega/2} + e^{-j\omega/2}}{2} \Big( (1+j)X(\omega) \ + \ (1-j)X(\omega-\pi) \Big) \\
& = \ e^{-j\omega/2} \ \cos(\omega/2) \ \Big( (1+j)X(\omega) \ + \ (1-j)X(\omega-\pi) \Big) \\
\end{align}$$
so the next question to ask is what does the spectrum of $X(\omega)$ look like for a bit stream $a[n]$ that looks like 00000000...? or 11111111...? or 01010101...? or 10101010...? or 00110011...? or 01100110...? and then what is the spectrum of the quadrature modulating signal $S(\omega)$ with those same bit sequences for $a[n]$?