Problem Description

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.

Intuition

To solve this problem, consider it as finding the current number of '1' characters in the string s and then determining the maximum potential increase in '1's by performing one optimized trade. The trade involves identifying consecutive blocks of '0's that are flanked by '1's, as swapping '0's to '1's in such blocks will increase the active sections.

A strategic approach employs two pointers to traverse the string s. We keep track of:

1. The sum of '1' sections already present.
2. The largest combined number of '0's from two adjacent sections that can be converted to '1's in a single trade. This is recorded in a variable mx.
3. The count of '0's in the previous block, stored in pre, so we can combine it optimally.

As we iterate through the string:

• For each continuous segment of the same character, calculate its length cnt.
• If the segment's character is '1', simply add the count cnt to the total active sections.
• If it's '0', check the potential increase in '1's by comparing and updating mx with pre + cnt. Then set pre to cnt to remember this for the next comparison.

Ultimately, the goal is to maximize the number of active sections, thus requiring finding these critical block combinations efficiently.

The solution computes this with a time complexity of O(n) and space complexity of O(1), making it efficient and optimal for linear string processing.

Solution Approach

The solution is based on a combination of greedy algorithm techniques and the two pointers pattern. Here's a breakdown of the approach:

1. Initialization:

   • Start by initializing several variables:
     • ans to keep track of the number of '1's counted so far.
     • pre to store the length of the previous block of '0's, initially set to a negative infinity to handle the start case.
     • mx to store the maximum number of contiguous '0' characters that can be converted by combining two segments, initially set to zero.

2. Iterate through the string:

   • Use a while loop with an index variable i to scan through the string s:
     • For each position, determine how many identical consecutive characters appear starting from position i, and denote this length as cur.
     • If the current character is '1', simply add cur to ans since this block's '1's are already active.
     • If the current character is '0', calculate a potential maximum by taking the sum pre + cur since these can potentially be converted to '1's via a trade. Update mx with the larger value between the current mx and pre + cur. Assign cur to pre for the next iteration to update correctly.

3. Calculate the result:

   • After the loop ends, the final result is calculated by adding mx to ans, representing the optimal scenario after performing the best possible trade.

Overall, the key idea here is to account for the '1's already present while seeking the best trade that maximizes additional '1' sections by combining contiguous blocks of '0's. The use of two pointers effectively manages bounded segments without unnecessary complexity, achieving outcomes in linear time O(n).

Example Walkthrough

Consider a small binary string s = "1001100".

1. Augmented String: Before starting, think of the string as augmented with '1' at both ends, forming t = "110011001".

2. Initialize Variables:

   • ans = 0 (to count current '1's).
   • pre = -∞ (to track the previous block of '0's).
   • mx = 0 (to track the maximum '0's that can be converted).

3. Iteration through s:

   • The first character is '1'. We count consecutive '1's: cur = 1. Add cur to ans: ans = 1.
   • The next character is '0'. We count consecutive '0's: cur = 1. Calculate mx as max(mx, pre + cur) = max(0, -∞ + 1) = 0. Set pre to cur: pre = 1.
   • The following character is '1'. Count consecutive '1's: cur = 2. Add cur to ans: ans = 3.
   • Next, the character is '0'. Count consecutive '0's: cur = 2. Now calculate mx as max(mx, pre + cur) = max(0, 1 + 2) = 3. Update pre to cur: pre = 2.

4. Calculate the Final Result:

   • After completing the loop, the potential maximum number of active sections is given by ans + mx = 3 + 3 = 6.

In this walk-through, the optimal trade is flipping the '0's between the two blocks of '1's: "1001100" -> "1111111", resulting in the maximum number of active '1' sections at 6.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n), where n is the length of the string s. This is because the code iterates through the entire string exactly once, with operations inside the loop taking constant time.

The space complexity is O(1), as the space used does not depend on the input size n. The algorithm uses a fixed amount of extra space for variables ans, i, pre, mx, cur, and j, which remain constant regardless of the input size.

Learn more about how to find time and space complexity quickly.