As pointed out by Snowball, the problem is inherently hard, see also this paper.
However, it can be done much faster in general than generating all the codewords.
I explain the idea for linear $[n,k]$ codes $C$ with $n = 2k$.
First, we construct two generator matrices $G_1 = (I\ A)$ and $G_2 = (B\ I)$ of $C$ where $I$ is the $(k\times k)$ unit matrix and $A,B$ are two $(k\times k)$ matrices. This is only possible if the positions $1,\ldots,k$ and $k+1,\ldots n$ form two information sets of $C$. If this is not the case, we have to find a suitable coordinate permutation first.
Now a codeword of weight $w$ has weight at most $\lfloor w/2\rfloor$ either in the first or in the second $n$ coordinates.
Thus, we can find all codewords of weight $\leq w$ among the codewords $x G_1 = (x, xA)$ and $x G_2 = (Bx, x)$, where $x$ is a vector of length $k$ and Hamming weight at most $\lfloor w/2\rfloor$.
So we can find the minimum distance $d$ of $C$ if we do this iteratively for all $w = 1,2,3,\ldots, d$.
In this way, you have to generate only a small fraction of all the codewords to find the minimum distance, and the idea can be generalized to any linear code. The first step then is to find a covering of the coordinates with information sets. However, the algorithm is still exponential, of course.
The webpage codetables.de is a valuable source for bounds on the best known minimum distances for fixed $q$, $n$ and $k$.
It is not true that $d$ depends solely on $n$ and $k$. For instance, let us take $n$ arbitrary and $k=1$, over the field $\mathbb{F}_2$. Then our code consists of simply two code words, $(0,\dots,0)$ and another codeword $c$. Note that if $c=(1,1,\dots,1)$, then the minimum distance is $n$. But if $c$ is, for instance, $(1,0,\dots,0)$, then the minimum distance is only $1$!
Using the terminology of an $[n,k,d]$ code gives you more information than you will usually have with just using an $[n,k]$ code. But occasionally if, for instance you do not know the minimum distance, then you can simply refer to it as a $[n,k]$ code.
Best Answer
Alright I figured this out, simply take the vector you want, lets call it $x$, then $d(x,x+u'')=w(x-(x+u''))=w(u'')=d(u,u')=d$.