Solving Codeforces Problem B1 Sub-RBS Easy | Generated by AI

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Question: How to solve the Codeforces problem “B1. Sub-RBS (Easy Version)”?

Answer:

This problem asks for the longest regular bracket subsequence t of a given regular bracket sequence s such that t is lexicographically better than s under a custom definition.

Let’s break it down step by step.

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1. Understanding “better than”

A bracket sequence t is better than s if:

1. s is a prefix of t and t ≠ s, or
2. At the first position where they differ, t[i] = '(' and s[i] = ')'.

Since t is a subsequence of s, the first case is impossible (a subsequence cannot be longer and still have s as a prefix).

So only the second rule matters:

At the first differing position, t must have '(' while s has ')'.

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2. Key observation

To be better than s, the subsequence t must:

• Match s for some prefix,
• Then skip a ')' in s and instead use a later '('.

That means: 👉 There must exist a position where s[i] = ')', but later in the string there is an unused '(' that we can pick.

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3. Strategy

We want the maximum length valid subsequence t.

To achieve this:

1. Remove a prefix of s ending at some ')'
2. From the remaining suffix, take the longest possible regular bracket subsequence

The optimal choice is:

• Remove the shortest prefix ending at a ')' such that the remaining suffix still contains a regular bracket sequence.

Why?

• Shorter removal → longer remaining subsequence
• Skipping a ')' immediately guarantees t is better than s

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4. When is it impossible?

If every prefix ending at ')' leaves an invalid suffix, then:

• No subsequence can be better than s
• Output -1

This happens when s is already lexicographically minimal, like:

• ()
• (())()

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5. Practical solution approach

For each test case:

1. Traverse the string from left to right.
2. Track balance of brackets.

3. Find the first position i where s[i] = ')' such that:

   • The suffix s[i+1 ... n] contains at least one valid RBS

4. Compute the length of the longest RBS subsequence in that suffix:

   • This is 2 × min(number of '(' , number of ')')
5. Output that length.
6. If no such position exists → output -1.

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6. Example walkthrough

Example 1

s = ()

• Only valid subsequence is itself
• No way to be better ✅ Output: -1

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Example 2

s = (()(()))

• Skip the ')' at position 3
• Remaining string can form ((()))
• Length = 6 ✅ Output: 6

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Example 3

s = (())()

• Any skipped ')' breaks validity of the suffix
• No better subsequence ✅ Output: -1

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7. Time complexity

• Each test case is processed in O(n)
• Total complexity across all test cases is O(Σn) ≤ 2 × 10⁵
• Fully acceptable within constraints

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8. Final takeaway

✔ You are not constructing the subsequence explicitly ✔ You are finding the best possible suffix after skipping one crucial ')' ✔ If such a suffix exists → compute its maximum RBS length ✔ Otherwise → -1

This greedy observation is the core idea behind solving the problem efficiently.

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