Let's say the N-Point DFT of $x[n]$ is $X(k)$.
Now from this question, I know how to compute the 2N-Point DFT of the same $x[n]$ in terms of $X(k)$
However, this idea doesn't translate well if I have to compute the 2N-Point DFT of $x_1[n]$, for example, where:
$$x_1[n] = \begin{cases} x[n], & \text{if } n \in [0, N-1]\\ x[n-N], & \text{if } n\in [N, 2N-1]\\ 0 & \text{else} \end{cases}$$
Or for example, $x_2[n]$, where:
$$x_2[n] = \begin{cases} x[n], & \text{if } n \in [0, N-1]\\ 0 & \text{else} \end{cases}$$
Although for $x_2[n]$, I can intuitively tell that $X(k) = X_2(k)$ since there is just zero padding past N. For the case $x_1[n]$ it is rather difficult to compute the 2N-Point DFT in terms of Even indexed and Odd indexed samples. So is there a more general way to compute the 2N-Point DFT in terms of N-Point DFT?