3499 - Maximize Active Section with Trade I.

Medium

You are given a binary string s of length n, where:

• '1' represents an active section.

• '0' represents an inactive section.

You can perform at most one trade to maximize the number of active sections in s. In a trade, you:

• Convert a contiguous block of '1's that is surrounded by '0's to all '0's.

• Afterward, convert a contiguous block of '0's that is surrounded by '1's to all '1's.

Return the maximum number of active sections in s after making the optimal trade.

Note: Treat s as if it is augmented with a '1' at both ends, forming t = '1' + s + '1'. The augmented '1's do not contribute to the final count.

Example 1:

Input: s = "01"

Output: 1

Explanation:

Because there is no block of '1's surrounded by '0's, no valid trade is possible. The maximum number of active sections is 1.

Example 2:

Input: s = "0100"

Output: 4

Explanation:

• String "0100" → Augmented to "101001".

• Choose "0100", convert <code>"10<ins> 1 </ins>001"</code> → <code>"1<ins> 0000 </ins>1"</code> → <code>"1<ins> 1111 </ins>1"</code>.

• The final string without augmentation is "1111". The maximum number of active sections is 4.

Example 3:

Input: s = "1000100"

Output: 7

Explanation:

• String "1000100" → Augmented to "110001001".

• Choose "000100", convert <code>"11000<ins> 1 </ins>001"</code> → <code>"11<ins> 000000 </ins>1"</code> → <code>"11<ins> 111111 </ins>1"</code>.

• The final string without augmentation is "1111111". The maximum number of active sections is 7.

Example 4:

Input: s = "01010"

Output: 4

Explanation:

• String "01010" → Augmented to "1010101".

• Choose "010", convert <code>"10<ins> 1 </ins>0101"</code> → <code>"1<ins> 000 </ins>101"</code> → <code>"1<ins> 111 </ins>101"</code>.

• The final string without augmentation is "11110". The maximum number of active sections is 4.

Constraints:

• <code>1 <= n == s.length <= 10⁵</code>

• s[i] is either '0' or '1'