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

1. Linear Time and Space Solution (Main focus)
2. 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

1. 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.
2. 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.