Blackslex and Password

The important move is to stop thinking about the string as one long line and instead split the positions by their remainder modulo $x$.

Two indices $i$ and $j$ satisfy

$$(j-i) \equiv 0 \pmod{x}$$

exactly when they are in the same residue class modulo $x$. In normal human words: they are separated by some multiple of $x$.

So the condition says:

• positions congruent to $1 \pmod{x}$ must all have pairwise different letters,
• positions congruent to $2 \pmod{x}$ must all have pairwise different letters,
• and so on.

But positions from different residue classes do not care about each other at all. They can reuse letters freely.

What makes a length invalid?

There are only $k$ available letters. In any one residue class, every character must be distinct, so that class can contain at most $k$ positions.

That means a string of length $n$ is valid exactly when every residue class modulo $x$ has size at most $k$.

There are $x$ residue classes, and each can hold at most $k$ positions. Therefore the maximum possible valid length is

$$kx.$$

For $n = kx$, we can fit exactly $k$ positions into each of the $x$ classes. For example, in each class we can use the letters $a,b,c,\dots$ independently. Reusing the same letter in different classes is allowed, because those positions are not constrained.

For $n = kx+1$, we have $kx+1$ positions distributed among only $x$ residue classes. By the pigeonhole principle, at least one residue class contains $k+1$ positions. That class would need $k+1$ pairwise distinct letters, but only $k$ letters exist. Impossible.

So the smallest bad length is simply

$$\boxed{kx+1}$$

No DP, no greedy, no string construction needed. Just modulo classes and pigeonholes. Very 800-rated, very blink-and-you-miss-it.