You are given a string s consisting only of the characters 'a' and 'b'.

You are allowed to repeatedly remove any substring where the number of 'a' characters is equal to the number of 'b' characters. After each removal, the remaining parts of the string are concatenated together without gaps.

Return an integer denoting the minimum possible length of the string after performing any number of such operations.

Example:

Suppose s = "aabbab". We can remove substrings whose count of 'a' equals the count of 'b'. For instance, we may remove "aabb" (2 a's and 2 b's), leaving "ab", and then remove "ab" (1 a and 1 b), leaving an empty string. The minimum possible length here is 0.

The key observation is that whenever two adjacent characters differ (one 'a' and one 'b'), they form a valid substring of equal counts and can be removed. By repeatedly removing such pairs, the string keeps shrinking until only one type of character remains. The leftover characters are simply the surplus of whichever character appears more frequently.

Therefore, if a is the count of 'a' and b is the count of 'b', the minimum possible length is the absolute difference between them, i.e. abs(a - b).

The core idea is to figure out which characters can actually be eliminated and which ones must remain no matter what we do.

Notice that whenever we have two adjacent characters that are different — that is, an 'a' next to a 'b' (or a 'b' next to an 'a') — they form a substring with exactly one 'a' and one 'b'. Since their counts are equal, this substring can be removed. After removing it, the characters that were on either side become adjacent, possibly creating new removable pairs.

This means we can keep "canceling out" pairs of different characters, much like matching opposite signs. Think of it as pushing characters together: every time an 'a' meets a 'b', they annihilate each other. So the question becomes — after canceling as many 'a'-'b' pairs as possible, what is left?

If we have a copies of 'a' and b copies of 'b', each removal consumes exactly one of each. We can keep removing until one type runs out. The character that appears more often will have some leftover copies that have nothing to pair with, and these can never be removed because any remaining substring would consist of only one type of character (which can never have equal counts unless it's empty).

So the final string consists purely of the surplus characters of whichever letter is more frequent. The number of these leftover characters is exactly abs(a - b), which is our minimum possible length. This also explains why the actual order of characters in s does not matter — only the counts of 'a' and 'b' determine the answer.

Pattern Learn more about Stack patterns.

Solution 1: Counting

Based on the intuition above, the order of characters doesn't matter — only how many 'a' and 'b' characters exist. Every 'a'-'b' pair can eventually be canceled out, so all that remains is the surplus of the more frequent character. This reduces the entire problem to a simple counting task.

The implementation follows these steps:

1. Count the 'a' characters. Use s.count("a") to get the total number of 'a' characters and store it in a.

2. Derive the 'b' count. Since the string contains only 'a' and 'b', the number of 'b' characters is simply the total length minus the count of 'a', i.e. b = len(s) - a. This avoids scanning the string a second time.

3. Return the surplus. The minimum possible length equals the absolute difference between the two counts, abs(a - b), which represents the leftover characters that can never be paired and removed.

No extra data structures or complex patterns are needed — just two counts and one subtraction.

The implementation is shown below:

class Solution:
    def minLengthAfterRemovals(self, s: str) -> int:
        a = s.count("a")
        b = len(s) - a
        return abs(a - b)

Complexity Analysis:

• Time Complexity: O(n), where n is the length of the string s. Counting the 'a' characters requires a single pass over the string.
• Space Complexity: O(1), since only a constant number of integer variables are used regardless of the input size.