What is the name of this class of (combinatorial?) problems?

I wouldn't tell you a name of such problem even if it exists, but I will show that the problem about maximum stretch of time is NP-hard. We can polynomially reduce SAT problem to the desired one in the following way.

Firstly I use an assumption that shoter string cannot come ouf of borders of the lonest one.

Given a formula $F$ of $m$ disjunctions on variables $x_1, x_2, \ldots, x_n$, we build the following set of strings. For each literal $x_1, \overline{x_1}, x_2, \ldots, \overline{x_n}$ we create a string of length $2m + 2n$ as follows:

• symbol at position $2i$ for $1 \le i \le m$ corresponds to the entry of corresponding literal into $i$th disjunction, i. e. is $1$ iff literal appears in $i$th disjunction;
• symbol at position $(2m + 2j - 1)$ for $1 \le j \le n$ is $1$ iff the string corresponds to either $x_j$ or $\overline{x_j};$
• all other symbols are $0$s.

One more string of length $2m + 2n + 1$ has $1$ at each odd position. It is easy to see that stretch of time is $2m + 2n + 1$ if and only if $F$ is satisfiable, because each even position $(2m + 2j)$ requires at least one of string corresponding to $x_j$ and $\overline{x_j}$ to be shifted by $1$ (i. e. requires this literal to be false), while each position $2i$ is a disjunction that requires at least one of srings corresponding to the literals of this disjunction to be shifted by $0$ (i. e. requires at least one true literal). Thus the problem about maximum stretch of time is NP-hard.

EDIT. Now we need to allow arbitrary shift of all string. For that purpose we add $4n(n + m)$ more symbols to each string. For $i$th string, $1 \le i \le 2n$ these symbols are $0$s except positions between $2i(n + m) + 1$ and $2(i + 1)(n + m)$, inclusive, where only $1$s are placed. The last string has $1$s at positions $2i(n + m) + 1$ for all $2 \le i \le 2n + 1$ and $0$s at all other positions.

It is not hard to see that we can get stretch of time $2(2n + 1)(n + m) + 1$ if and only if $F$ is satisfiable. And shifts of all strings should be $0$ or $1$ to make such stretch of time.

Example. Let $$F = (x_1 \lor x_2) \land (x_1 \lor \overline{x_2}) \land (\overline{x_1} \lor x_2) \land (\overline{x_1} \lor \overline{x_2}).$$

Then corresponding set of strings is the following:

$x_1\colon s_1 =$ 01010000 1000 111111111111 000000000000 000000000000 000000000000
$\overline{x_1}\colon s_2 =$ 00000101 1000 000000000000 111111111111 000000000000 000000000000
$x_2\colon s_3 =$ 01000100 0010 000000000000 000000000000 111111111111 000000000000
$\overline{x_2}\colon s_4 =$ 00010001 0010 000000000000 000000000000 000000000000 111111111111
and $s_5 =$ 10101010 1010 100000000000 100000000000 100000000000 100000000000 1

It is not hard to see that we cannot get stretch of time $61$ therefore $F$ is not satisfiable.