CF 2178B - Impost or Sus

Rating: 900
Tags: dp, greedy, implementation, strings
Solve time: 2m 35s
Verified: no

Solution

Problem Understanding

We start with a string consisting only of s and u.

We may repeatedly choose any position containing u and change it into s. We are not allowed to change an s back into u.

The final string must satisfy two conditions. First, it must contain at least two occurrences of s. Second, every remaining u must sit exactly halfway between its two nearest s characters.

The task is to find the minimum number of u → s operations required.

The total length of all strings across the test cases is at most 2 · 10^5. Any solution that examines each character only a constant number of times is easily fast enough. Quadratic approaches are ruled out because a single test could already contain 200000 characters.

The tricky part is understanding what a valid final string actually looks like.

Consider two consecutive s characters in the final string.

If their distance is 1, they form "ss" and there is no u between them.

If their distance is 2, they form "sus", and the middle position is perfectly centered.

If the distance is larger than 2, there are multiple positions between them. Only the exact midpoint would be equally distant from both s, so every other position would violate the condition.

This means that in any suspicious string:

every u is surrounded by s on both sides,
no two u characters are adjacent,
the first and last characters cannot be u.

For example, "suss" is valid because the only u is between two s characters.

For example, "suus" is invalid because the two middle u characters cannot both be centered between the same pair of s.

A common mistake is to focus on distances between existing s characters. Since we are allowed to create new s characters anywhere by converting u, the important question is not where the current s characters are, but which u characters we choose to keep.

Consider:

uuuu

A greedy idea might try to place s only at the ends. That gives "suus", which is still invalid. The optimal answer is 3, producing something like "suss".

Another easy-to-miss case is:

sus

The answer is 0. The string already satisfies all requirements.

Finally:

uuu

Only the middle u can remain. Both end positions must become s, so the answer is 2.

Approaches

A brute-force solution would consider every subset of u positions to convert into s. For each choice, we could check whether the resulting string is suspicious.

If a string contains k occurrences of u, this requires examining 2^k possibilities. With k potentially near 200000, this is completely infeasible.

To do better, we need to characterize which u characters are allowed to survive.

From the structural observation above, every surviving u must satisfy two properties.

First, it cannot be at either end of the string.

Second, it cannot be adjacent to another surviving u.

Nothing else matters.

If a u is internal and all neighboring positions become s, then it automatically sits between two s characters at distance 1 on both sides.

This transforms the problem into a maximum-keeping problem.

We want to keep as many original u characters as possible, subject to:

kept u positions are not at the ends,
no two kept u positions are adjacent.

Every u that is not kept must be converted into s.

So:

answer = total_u - maximum_keepable_u

Finding the maximum number of non-adjacent kept positions is a standard path DP.

Let:

dp0 = best number of kept u so far when the previous position is not a kept u,
dp1 = best number of kept u so far when the previous position is a kept u.

We scan left to right and update these states in constant time per character.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^k · n)	O(n)	Too slow
Optimal DP	O(n)	O(1)	Accepted

Algorithm Walkthrough

Count the number of u characters in the string.
Maintain two DP states.

dp0 stores the maximum number of kept u characters so far when the previous position is not kept.

dp1 stores the maximum number of kept u characters so far when the previous position is kept. 3. Process the string from left to right. 4. If the current character is s, it cannot be kept as a u.

Merge both previous states into the new dp0. 5. If the current character is u, we have two choices.

Convert it to s.

In that case, the previous state does not matter, and we transition into dp0. 6. If the current character is an internal position, meaning it is neither the first nor the last character, we may keep it as u.

To keep it, the previous position must not already be a kept u, so the transition comes only from dp0. 7. After processing all positions, the maximum number of keepable u characters is max(dp0, dp1). 8. The answer is:

total_u - maximum_keepable_u

Why it works

A suspicious string is completely determined by which original u characters remain unchanged.

A remaining u mu