How do you solve Maximum Score From Removing Substrings on LeetCode?

Maximum Score From Removing Substrings (LeetCode #1717) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Removing 'ab' or 'ba' can be done in any order, but the score gained from each operation differs.

What is the time complexity of Maximum Score From Removing Substrings?

The optimal solution for Maximum Score From Removing Substrings runs in O(n) time and uses O(n) space. This complexity is due to a single pass through the string, where each character is pushed or popped from the stack at most once.

What is the brute force solution for Maximum Score From Removing Substrings?

The brute force approach for Maximum Score From Removing Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible combination of removing 'ab' and 'ba' substrings from the string. This method guarantees that we explore all possibilities but is inefficient due to its high time complexity. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Score From Removing Substrings?

Maximum Score From Removing Substrings uses the following concepts: String, Stack, Greedy. The recommended patterns to study are: Stack, Greedy.

What are the interview tips for Maximum Score From Removing Substrings?

Always think about the data structures that can help you manage the problem effectively. Explain your thought process clearly, especially when transitioning from a brute-force to an optimal solution. Practice similar problems to get comfortable with using stacks and greedy algorithms.

What are common mistakes when solving Maximum Score From Removing Substrings?

Not considering the order of operations and how it affects the score. Failing to use a stack or similar data structure to manage the characters efficiently.

#1717

Maximum Score From Removing Substrings

Medium

StringStackGreedyStackGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a stack to efficiently manage the removal of 'ab' and 'ba' substrings while keeping track of the score. This approach ensures that we only traverse the string once, making it much faster.

⚙️

Algorithm

5 steps

1Step 1: Initialize a stack to keep track of characters and a score variable to 0.
2Step 2: Iterate through each character in the string.
3Step 3: For each character, check the top of the stack. If it forms 'ab' or 'ba' with the current character, pop from the stack and update the score accordingly.
4Step 4: If it doesn't form a substring, push the current character onto the stack.
5Step 5: Continue until all characters are processed.

solution.py16 lines

1# Full working Python code
2
3def maxScore(s: str, x: int, y: int) -> int:
4    stack = []
5    score = 0
6    for char in s:
7        if stack and ((stack[-1] == 'a' and char == 'b') or (stack[-1] == 'b' and char == 'a')):
8            if stack[-1] == 'a':
9                stack.pop()
10                score += x
11            else:
12                stack.pop()
13                score += y
14        else:
15            stack.append(char)
16    return score

ℹ

Complexity note: This complexity is due to a single pass through the string, where each character is pushed or popped from the stack at most once.

1Removing 'ab' or 'ba' can be done in any order, but the score gained from each operation differs.
2Using a stack allows us to efficiently manage the removal of substrings while keeping track of the score.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.