Problem Description

You have a binary string s consisting of '0's and '1's, where '1' represents an active section and '0' represents an inactive section.

You can perform at most one trade operation to maximize the number of active sections. A trade consists of two steps:

1. First, select a contiguous block of '1's that is surrounded by '0's on both sides and convert all those '1's to '0's
2. Then, select a contiguous block of '0's that is surrounded by '1's on both sides and convert all those '0's to '1's

The goal is to find the maximum possible number of active sections (total count of '1's) after performing the optimal trade.

An important note: The string is treated as if it has an imaginary '1' added at both the beginning and end (forming t = '1' + s + '1'). These imaginary '1's help define what blocks can be traded but don't count toward the final answer.

For example, if s = "0110100", it's treated as if surrounded by '1's: "10110100". The block of '0's at the beginning and end can be converted to '1's since they're surrounded by '1's (including the imaginary ones). Similarly, the middle blocks of '1's can be converted to '0's since they're surrounded by '0's.

The problem essentially asks you to find the best trade that maximizes the total count of '1's in the final string.

Intuition

When we perform a trade, we're removing some '1's and adding some '0's that get converted to '1's. The net change in active sections is: (number of '0's we convert to '1's) - (number of '1's we convert to '0's).

To maximize the final count, we want to maximize this net change. Since we can only trade blocks that are "surrounded" (a block of '1's surrounded by '0's, and a block of '0's surrounded by '1's), we need to think about which blocks are available for trading.

A key observation is that when we remove a block of '1's that's surrounded by '0's, we're essentially merging two adjacent blocks of '0's into one larger block. This merged block of '0's can then be converted to '1's in the second part of the trade.

For example, in "001110000111", if we remove the first block of '1's ("111"), we get "000000000111". The first 9 '0's are now one contiguous block that can be converted to '1's, giving us "111111111111".

This means the best trade is to:

1. Remove a block of '1's to merge two adjacent '0' blocks
2. Convert the merged '0' block to '1's

The gain from this trade is the sum of the two adjacent '0' blocks minus the '1' block between them. But since we're converting all the '0's to '1's anyway, we can simplify: the maximum gain is just the sum of two adjacent '0' blocks.

Therefore, our strategy is to:

• Count all existing '1's in the string (these will remain unless we trade them away)
• Find the maximum sum of two adjacent '0' blocks (this represents the best possible gain from a trade)
• Add these two values together

The solution iterates through the string, tracking consecutive segments. When we encounter '1' segments, we add their length to our answer. When we encounter '0' segments, we keep track of the maximum sum of the current '0' segment plus the previous '0' segment, which represents the best possible trade we could make.

Solution Approach

The implementation uses a two-pointer technique to traverse the string and identify consecutive segments of the same character. Here's how the algorithm works:

Variables Used:

• ans: Accumulates the total count of '1's in the original string
• mx: Tracks the maximum sum of two adjacent '0' segments (best possible trade gain)
• pre: Stores the length of the previous '0' segment
• i and j: Two pointers to identify consecutive segments

Algorithm Steps:

1. Initialize variables: Start with ans = 0, mx = 0, and pre = -inf (negative infinity ensures the first '0' segment doesn't incorrectly contribute to mx).

2. Segment identification: Use pointer i to mark the start of each segment. Move pointer j forward while s[j] == s[i] to find the end of the current consecutive segment. The length of this segment is cur = j - i.

3. Process each segment:

   • If the segment contains '1's: Add cur to ans (these are existing active sections)
   • If the segment contains '0's:
     • Update mx = max(mx, pre + cur) to track the best sum of two adjacent '0' segments
     • Update pre = cur for the next iteration

4. Move to next segment: Set i = j to start processing the next segment.

5. Final calculation: After processing all segments, add mx to ans. This represents the original '1' count plus the maximum gain from the optimal trade.

Why this works:

• By tracking pre + cur for consecutive '0' segments, we're effectively finding which two adjacent '0' blocks (separated by a '1' block) would give us the maximum gain if we traded them.
• The pre variable maintains a sliding window of opportunity - it always holds the previous '0' segment's length, allowing us to calculate the sum with the current '0' segment.
• When we encounter a '1' segment between two '0' segments, we don't reset pre, which correctly models that these two '0' segments are adjacent and can be merged through a trade.

Time Complexity: O(n) where n is the length of string s, as we traverse the string once.

Space Complexity: O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through the solution with s = "00111000".

Step 1: Add imaginary boundaries We conceptually treat the string as "1001110001" (with imaginary '1's at both ends), though we don't modify the actual string.

Step 2: Initialize variables

• ans = 0 (count of original '1's)
• mx = 0 (best trade gain)
• pre = -inf (previous '0' segment length)
• i = 0, j = 0

Step 3: Process segments

Segment 1: "00" (positions 0-1)

• i = 0, move j while s[j] == '0': j = 2
• Length cur = 2 - 0 = 2
• Since it's '0's: mx = max(0, -inf + 2) = 0, pre = 2
• Move i = 2

Segment 2: "111" (positions 2-4)

• i = 2, move j while s[j] == '1': j = 5
• Length cur = 5 - 2 = 3
• Since it's '1's: ans = 0 + 3 = 3
• Move i = 5

Segment 3: "000" (positions 5-7)

• i = 5, move j while s[j] == '0': j = 8
• Length cur = 8 - 5 = 3
• Since it's '0's: mx = max(0, 2 + 3) = 5, pre = 3
• Move i = 8

Step 4: Calculate final result

• Original '1's: ans = 3
• Best trade gain: mx = 5
• Final answer: 3 + 5 = 8

What happened? The algorithm identified that we can trade the middle "111" block (which is surrounded by '0's) to merge the two '0' blocks into one block of 5 '0's. Then we convert those 5 '0's to '1's (they're surrounded by the imaginary '1's). The net effect: we lose 3 '1's but gain 5 '1's, for a total of 8 '1's in the final string "11111111".

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the input string s.

The algorithm uses a single while loop that iterates through the string exactly once. In each iteration:

• The outer while loop moves the pointer i forward
• The inner while loop moves pointer j to find consecutive identical characters
• Each character in the string is visited exactly once by the combination of both loops
• All operations inside the loops (comparisons, max calculations, additions) are O(1)

Since each character is processed exactly once, the overall time complexity is linear.

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space:

• Variables n, ans, i, j for iteration and counting
• Variables pre, mx, cur for tracking previous segment, maximum gain, and current segment length
• All variables store single integer values regardless of input size

No additional data structures that scale with input size are used, resulting in constant space complexity.

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

Common Pitfalls

1. Misunderstanding the Trade Mechanism

Pitfall: A common mistake is thinking that we can trade ANY block of '1's for ANY block of '0's, rather than understanding that we can only trade blocks that are "surrounded" by the opposite character.

Example of the mistake: For string s = "1100011", someone might think:

• Trade the first two '1's (giving up 2)
• Convert the three '0's to '1's (gaining 3)
• Net gain: +1

But this is incorrect because the first two '1's are NOT surrounded by '0's on both sides (there's no '0' before them, even with the imaginary '1' prefix).

Solution: Remember that with the imaginary '1's added (t = '1' + s + '1'):

• Only '1' blocks that have '0's on BOTH sides can be traded away
• Only '0' blocks that have '1's on BOTH sides can be converted to '1's

2. Incorrect Initialization of previous_zero_segment

Pitfall: Initializing previous_zero_segment to 0 instead of negative infinity.

# WRONG
previous_zero_segment = 0  # This would incorrectly count the first segment

# CORRECT
previous_zero_segment = float('-inf')

Why it matters: If the string starts with '0's (like "00111"), and we initialize to 0, the first '0' segment would incorrectly contribute to max_zero_gain as if there was a previous '0' segment to pair with.

3. Forgetting Adjacent '0' Segments Must Be Separated by '1's

Pitfall: Not understanding that for two '0' segments to be part of the same trade, they must be separated by at least one '1' segment that gets traded away.

Example: In string "0011100111":

• The two '0' segments at positions [0-1] and [5-5] can be part of the same trade
• They would convert the '1's at positions [2-4] to '0's
• Then convert all positions [0-5] to '1's

The algorithm handles this correctly by maintaining previous_zero_segment across '1' segments, but it's easy to misunderstand why.

4. Off-by-One Errors in Segment Calculation

Pitfall: Incorrectly calculating segment length or boundaries.

# WRONG - might cause index out of bounds
while segment_end <= n and s[segment_end] == s[index]:
    segment_end += 1

# CORRECT
while segment_end < n and s[segment_end] == s[index]:
    segment_end += 1

5. Not Handling Edge Cases

Pitfall: Failing to consider special cases like:

• Empty string or single character string
• String with all '0's or all '1's
• String with no valid trades possible

Solution: The algorithm naturally handles these cases:

• All '1's: max_zero_gain stays 0, returns original count
• All '0's: Returns the length of the string (all can be converted due to imaginary '1's at boundaries)
• No trades: max_zero_gain represents the best possible trade (could be 0)