Maximum Score from Removing Substrings

Problem Statement

Given a string and two hardcoded substrings of size two (ab and ba)
Scores: ab = 4, ba = 5
Remove substrings to maximize the total score
Return only the score, not the modified string

Solutions Overview

Linear Time and Space Solution (Main focus)
Optimized Space Complexity (Brief discussion)

Key Observations

Both substrings ab and ba are of size two.
Removing a pair can introduce new pairs.
Problem has a greedy nature: prioritize removing higher-scoring pairs.

Detailed Solution: Linear Time and Space

Phase 1: Removing Pairs with Higher Scores

Iterate through the string with a stack.
Track pairs and remove them while keeping track of their corresponding scores.
Ensure stack-based approach to efficiently remove pairs.

Stack Data Structure

Use a stack to keep track of characters while iterating.
Pop from the stack when there’s a match (i.e., form a pair).
Add to the stack when no pair is formed.
Update the score accordingly.

Phase 2: Removing Remaining Pairs

Second pass through the updated string to remove the other type of substrings.
Ensure no pairs from the first type resurface.

Proof by Contradiction

Ensures that after the first phase, no previous type pairs (ab) will exist to conflict during the second phase.
Use logic checks to confirm that all pairs are managed correctly so that no additional pairs are introduced erroneously.

Implementation Steps

Helper Function: remove_pairs
- Takes in the string, pair type, and score.
- Uses a stack to manage additions and removals of characters.
- Updates the input string and score tally.
Main Function
- Initialize result to zero.
- Use remove_pairs twice: first for the higher score pair, then the other.
- Return the total score.

Python Code Example

class Solution:
    def maximumGain(self, s: str, x: int, y: int) -> int:
        def remove_pairs(s, pair, score):
            stack = []
            total_score = 0
            for char in s:
                if stack and stack[-1] == pair[0] and char == pair[1]:
                    stack.pop()
                    total_score += score
                else:
                    stack.append(char)
            return ''.join(stack), total_score

        first_pair, second_pair = ('ab', 'ba') if x > y else ('ba', 'ab')
        first_score, second_score = (x, y) if x > y else (y, x)

        s, first_total = remove_pairs(s, first_pair, first_score)
        s, second_total = remove_pairs(s, second_pair, second_score)

        return first_total + second_total

Function remove_pairs: Removes pair from the string and returns the modified string and total score.
Main function uses remove_pairs for both pairs based on their scores.

Conceptual Discussion: Constant Space Solution

Hypothetical Implementation: If the input was an array instead of a string.
Technique: Use two pointers to shift elements in place and remove pairs.
Managing Spaces: Overwrite elements directly in the array to avoid extra space usage.

Summary

Efficient linear solution using stack-based approach.
Greedy methodology to maximize score by removing higher-scoring pairs first.
Proof by contradiction ensures correctness of the strategy.