A string that takes a DFA from any state to a given state?
It turns out that there exists a string that takes all states to the same state iff for each pair of states, there exists a string that takes both states to the same state.

Using this fact, I had success with the following algorithm:


Generate a new DFA from the Cartesian product of the old DFA with itself. (The new DFA contains $\left(a,b\right) \overset{x}{\rightarrow} \left(c,d\right)$ iff the old DFA contains $a \overset{x}{\rightarrow} c$ and $b \overset{x}{\rightarrow} d$, where $x$ is an input symbol and $a$, $b$, $c$, and $d$ are states.)
For each state $\left(a,b\right), a \neq b$ in the new DFA, search for a path from $\left(a,b\right)$ to a state of the form $\left(c,c\right)$. (If found, then that means there is a string that takes the old DFA from $a$ to $c$ and from $b$ to $c$.)
There is a reset sequence for the old DFA iff every search in the above step yields such a path.


More info:

https://www.ics.uci.edu/~eppstein/pubs/Epp-SJC-90.pdf