New Year String

The definition looks a little weird, so simplify it first.

A string is a New Year string if:

• it contains 2026, or
• it does not contain 2025.

That means the only strings that fail are strings that contain 2025 while also not containing 2026. So we can make the string valid in either of two ways:

1. create at least one 2026,
2. destroy every 2025.

The minimum answer is the cheaper of those two. No need to overcook this. It is an 800, not a villain origin story.

Making 2026

If we want the final string to contain 2026, we only need to choose one window of length $4$ and change that window into 2026.

For a window starting at index $i$, the cost is:

$$[s_i \ne 2] + [s_{i+1} \ne 0] + [s_{i+2} \ne 2] + [s_{i+3} \ne 6]$$

Take the minimum over all windows.

Everything outside that window can stay unchanged, because the condition only asks for at least one occurrence of 2026.

Removing all 2025

Now we need the minimum number of edits so the final string avoids 2025 as a substring.

Since $n \le 20$, many things would work, but the cleanest robust method is a pattern-avoidance DP.

Let the forbidden pattern be:

$$p = \texttt{2025}$$

Define:

$$dp[i][k]$$

as the minimum number of changes after building the first $i$ characters, where $k$ is the length of the longest suffix of the built string that is also a prefix of 2025.

So $k$ can be $0,1,2,3$. We never allow $k=4$, because that means we just formed the forbidden substring 2025.

For each position $i$, we try every possible replacement character $c \in \{0,2,5,6\}$. The cost of choosing $c$ is $0$ if $c=s_i$, otherwise $1$.

Then we update the matched prefix length. For example, if our current suffix state is represented by the first $k$ characters of 2025, we append $c$, and compute the longest suffix of that new string that is also a prefix of 2025.

If that new length is $4$, skip it. That transition creates 2025, which is exactly the crap we are trying to avoid.

At the end, the cost to remove all 2025 is:

$$\min_{k=0}^{3} dp[n][k]$$

Why this is correct

For the 2026 plan, any valid final string containing 2026 has some occurrence of 2026. Looking at that occurrence gives one window whose mismatch count is at most the total number of edits. Conversely, changing that cheapest window is enough. So the window minimum is exact.

For the no-2025 plan, the DP stores exactly the information needed to know whether the next character completes 2025: the longest suffix that could still become the forbidden pattern. This is the standard finite automaton idea, just tiny enough to write directly. Every edited string avoiding 2025 corresponds to one valid DP path, and every valid DP path builds a string avoiding 2025. Therefore the DP minimum is exact.

Finally, a string is valid if it satisfies either plan, so the answer is:

$$\min(\text{minimum cost to contain `2026`},\ \text{minimum cost to avoid `2025`})$$

Complexity

There are $n$ positions, $4$ states, and $4$ possible characters per transition. The transition itself checks strings of length at most $4$, so this is basically constant-time per test case.

Formally:

$$O(n \cdot 4 \cdot 4 \cdot 4)$$

which is hilariously fine for $n \le 20$ and $t \le 10^4$.