How do you solve Minimum String Length After Removing Substrings on LeetCode?

Minimum String Length After Removing Substrings (LeetCode #2696) 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: Using a stack allows for efficient removal of adjacent substrings in constant time.

What is the brute force solution for Minimum String Length After Removing Substrings?

The brute force approach for Minimum String Length After Removing Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly scanning the string to find and remove occurrences of the substrings 'AB' and 'CD'. This is done until no more occurrences are found, resulting in a potentially reduced string length. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum String Length After Removing Substrings?

Minimum String Length After Removing Substrings uses the following concepts: String, Stack, Simulation. The recommended patterns to study are: Stack, Simulation.

What are the interview tips for Minimum String Length After Removing Substrings?

Always think about edge cases, such as strings that do not contain 'AB' or 'CD'. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Practice similar problems to recognize patterns, such as using stacks for substring removal.

What are common mistakes when solving Minimum String Length After Removing Substrings?

Not considering that removing one substring can create new substrings. Failing to optimize the solution from a brute force approach to a more efficient one.

#2696

Minimum String Length After Removing Substrings

Easy

StringStackSimulationStackSimulation

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 approach uses a stack to efficiently manage the removal of substrings. By pushing characters onto the stack and checking for 'AB' or 'CD' at the top, we can remove them in constant time, leading to a linear time complexity.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty stack.
2Step 2: Iterate through each character in the string.
3Step 3: Push the character onto the stack. If the last two characters in the stack form 'AB' or 'CD', pop them off.
4Step 4: Continue this until all characters are processed.
5Step 5: The remaining characters in the stack represent the final string.

solution.py8 lines

1def min_length_after_removal(s):
2    stack = []
3    for char in s:
4        stack.append(char)
5        if len(stack) >= 2 and (stack[-1] == 'B' and stack[-2] == 'A' or stack[-1] == 'D' and stack[-2] == 'C'):
6            stack.pop()
7            stack.pop()
8    return len(stack)

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once, and each character is pushed and popped from the stack at most once. The space complexity is O(n) due to the stack storing the characters.

1Using a stack allows for efficient removal of adjacent substrings in constant time.
2The problem can be visualized as a series of operations that modify the string dynamically.

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