Problem Description

You are given a string s and two integers x and y. You can perform two types of operations any number of times:

1. Remove substring "ab" and gain x points.

   • For example, when removing "ab" from "cabxbae", it becomes "cxbae".

2. Remove substring "ba" and gain y points.

   • For example, when removing "ba" from "cabxbae", it becomes "cabxe".

Your goal is to return the maximum points you can gain after applying these operations on string s.

The key insight is that you can perform these operations multiple times in any order. Each time you remove either "ab" or "ba", you earn the corresponding points (x or y). The challenge is to determine the optimal sequence of removals to maximize your total score.

For example, if you have the string "aabb", you could:

• Remove "ab" first to get "ab" and earn x points
• Then remove the remaining "ab" to earn another x points
• Total: 2x points

Or you could manipulate the order differently if that yields more points based on the values of x and y.

The problem requires finding the strategy that maximizes the total points earned from all possible removal operations.

Intuition

The first key observation is that we should prioritize removing the substring that gives us more points. If x > y, we should try to remove "ab" as much as possible before removing "ba". If y > x, we should do the opposite.

Why does this greedy approach work? Consider what happens when we have a mix of 'a's and 'b's. Each removal operation consumes exactly one 'a' and one 'b'. No matter what order we perform the operations, the total number of operations we can perform is fixed - it's min(count_of_a, count_of_b).

Since the total number of operations is fixed, we want to maximize the value we get from these operations. The greedy strategy is to perform the higher-value operations first, and only perform the lower-value operations with whatever characters are left.

Another crucial insight is that characters other than 'a' and 'b' act as barriers. They split the string into independent segments. For example, in "aabcbba", the 'c' divides it into "aab" and "bba", which we can process separately.

The clever part of the solution is how it processes the string in a single pass. As we traverse:

• When we see an 'a', we hold onto it (increment counter) hoping to pair it with a future 'b' for the higher-value operation
• When we see a 'b', we check if we have any 'a's waiting. If yes, we immediately form the high-value pair. If no, we hold onto this 'b'
• When we see any other character, we know we've reached the end of a segment. Any unpaired 'a's and 'b's in this segment can only form the lower-value pairs, so we calculate min(cnt_a, cnt_b) * lower_value

This approach ensures we always prioritize the higher-value operations while processing the string in linear time.

Learn more about Stack and Greedy patterns.

Solution Approach

The implementation follows a greedy strategy with a single-pass traversal:

Step 1: Normalize the problem

a, b = "a", "b"
if x < y:
    x, y = y, x
    a, b = b, a

We ensure that x ≥ y by swapping if necessary. This means we always prioritize removing the pattern stored in variables a and b (which forms "ab" or "ba" depending on the swap).

Step 2: Initialize counters

ans = cnt1 = cnt2 = 0

• ans: accumulates the total score
• cnt1: counts unpaired instances of the first character (a)
• cnt2: counts unpaired instances of the second character (b)

Step 3: Process each character For each character c in the string:

• Case 1: c == a (first character of high-value pattern)

  cnt1 += 1

  We increment cnt1 to track this character, hoping to pair it with a future b.

• Case 2: c == b (second character of high-value pattern)

  if cnt1:
      ans += x
      cnt1 -= 1
  else:
      cnt2 += 1

  If we have unpaired as (cnt1 > 0), we can form the high-value pattern immediately, earning x points. Otherwise, we store this b in cnt2 for potential low-value pairing later.

• Case 3: c is neither a nor b

  ans += min(cnt1, cnt2) * y
  cnt1 = cnt2 = 0

  This character acts as a segment boundary. Any remaining unpaired as and bs can only form low-value patterns. We add min(cnt1, cnt2) * y to the score and reset both counters.

Step 4: Handle remaining characters

ans += min(cnt1, cnt2) * y

After processing all characters, we handle any remaining unpaired characters at the end of the string, treating them as low-value patterns.

This algorithm runs in O(n) time with O(1) extra space, where n is the length of the string. The key insight is that by always greedily forming high-value patterns first during traversal, we ensure maximum score without needing to actually modify the string or use additional data structures like stacks.

Example Walkthrough

Let's walk through the solution with string s = "aabbaxyab", x = 4, and y = 5.

Since y > x, we first swap values: x = 5, y = 4, and swap patterns: a = "b", b = "a". Now we prioritize removing "ba" (worth 5 points) over "ab" (worth 4 points).

Initial state: ans = 0, cnt1 = 0, cnt2 = 0

Processing each character:

1. 'a' (index 0): This is our b character (after swap)

   • No cnt1, so cnt2 = 1
   • State: cnt1 = 0, cnt2 = 1, ans = 0

2. 'a' (index 1): Another b character

   • No cnt1, so cnt2 = 2
   • State: cnt1 = 0, cnt2 = 2, ans = 0

3. 'b' (index 2): This is our a character (after swap)

   • cnt1 = 1
   • State: cnt1 = 1, cnt2 = 2, ans = 0

4. 'b' (index 3): Another a character

   • cnt1 = 2
   • State: cnt1 = 2, cnt2 = 2, ans = 0

5. 'a' (index 4): This is our b character

   • We have cnt1 > 0, so we can form "ba"!
   • ans += 5, cnt1 -= 1
   • State: cnt1 = 1, cnt2 = 2, ans = 5

6. 'x' (index 5): Neither 'a' nor 'b' - segment boundary!

   • Form low-value pairs: min(1, 2) * 4 = 4
   • ans += 4, reset counters
   • State: cnt1 = 0, cnt2 = 0, ans = 9

7. 'y' (index 6): Neither 'a' nor 'b' - segment boundary!

   • min(0, 0) * 4 = 0 (nothing to pair)
   • State: cnt1 = 0, cnt2 = 0, ans = 9

8. 'a' (index 7): This is our b character

   • No cnt1, so cnt2 = 1
   • State: cnt1 = 0, cnt2 = 1, ans = 9

9. 'b' (index 8): This is our a character

   • cnt1 = 1
   • State: cnt1 = 1, cnt2 = 1, ans = 9

Final step: Handle remaining characters

• min(1, 1) * 4 = 4
• ans = 9 + 4 = 13

Result: Maximum points = 13

The algorithm successfully identified that we could form one "ba" (5 points) in the first segment, one "ab" (4 points) from the remaining characters in that segment, and one "ab" (4 points) in the last segment, giving us a total of 13 points.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of string s. The algorithm iterates through the string exactly once using a single for loop that processes each character. Each operation inside the loop (comparisons, arithmetic operations, and variable assignments) takes constant time O(1). Therefore, the overall time complexity is linear with respect to the input size.

The space complexity is O(1). The algorithm uses only a fixed number of variables (a, b, ans, cnt1, cnt2, x, y) regardless of the input size. No additional data structures that grow with the input are created. The space usage remains constant throughout the execution of the algorithm.

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

Common Pitfalls

1. Not Handling the Greedy Order Correctly

Pitfall: A common mistake is trying to remove both "ab" and "ba" patterns simultaneously without considering which one should be prioritized. Simply scanning for both patterns in a single pass will lead to suboptimal results.

Why it happens: Consider the string "aabb" with x=4 and y=1. If you naively remove patterns as you encounter them without prioritization, you might remove "ba" first (middle characters), leaving you with "ab" for a total of 5 points. However, removing two "ab" patterns would give you 8 points.

Solution: Always prioritize removing the higher-value pattern first. The code handles this by normalizing the problem - ensuring x ≥ y through swapping, then greedily forming the higher-value pattern during traversal.

2. Using String Manipulation Instead of Counters

Pitfall: Actually removing substrings from the string during processing, like using s = s.replace("ab", "") repeatedly.

Why it's problematic:

• String manipulation in Python creates new strings (O(n) per operation)
• Multiple removals can lead to O(n²) time complexity
• After each removal, you need to rescan the string for new patterns that may have formed

Example problematic code:

# DON'T DO THIS
while "ab" in s or "ba" in s:
    if x >= y and "ab" in s:
        s = s.replace("ab", "", 1)
        score += x
    elif "ba" in s:
        s = s.replace("ba", "", 1)
        score += y

Solution: Use counters to track unpaired characters instead of modifying the string. This maintains O(n) time complexity with a single pass.

3. Forgetting to Process Remaining Characters

Pitfall: Only processing characters when encountering non-a/b characters as segment boundaries, but forgetting to handle leftover characters at the end of the string.

Example: For string "abab", if you only process at segment boundaries (non-a/b characters), you'd miss the final characters and lose potential points.

Solution: Always add a final processing step after the loop:

# Don't forget this line!
total_score += min(first_char_count, second_char_count) * y

4. Incorrect Swap Logic

Pitfall: When x < y, only swapping the values but not the character mappings, or vice versa.

Incorrect approach:

# Wrong - only swapping values
if x < y:
    x, y = y, x
    # But still looking for "ab" first instead of "ba"

Solution: Swap both the point values AND the character order to maintain consistency:

if x < y:
    x, y = y, x
    first_char, second_char = "b", "a"  # Also swap the pattern