What is the time complexity of Minimum String Length After Removing Substrings?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum String Length After Removing Substrings?

This problem uses String, Stack, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum String Length After Removing Substrings?

Insight 1: Using a stack (or similar data structure) to efficiently track characters and identify "AB" or "CD" subsequences is crucial for avoiding redundant checks. Insight 2: The problem can be solved iteratively by repeatedly removing "AB" and "CD" until no more such substrings are found. A stack efficiently manages this process by only considering adjacent pairs of characters.

Which companies ask LeetCode 2696. Minimum String Length After Removing Substrings in interviews?

This problem is frequently asked at J.P. Morgan, Yelp.

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Minimum String Length After Removing Substrings - LeetCode 2696 Solution

Minimum String Length After Removing Substrings - Complete Solution Guide

Minimum String Length After Removing Substrings is LeetCode problem 2696, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a string s consisting only of uppercase English letters. You can apply some operations to this string where, in one operation, you can remove any occurrence of one of the substrings "AB" or "CD" from s . Return the minimum possible length of the resulting string that you can obtain . Note that the string concatenates after removing the substring and could produce new "AB" or "CD" substrings. Example 1: Input: s = "ABFCACDB" Output: 2 Explanation: We can do the following operations:

Detailed Explanation

The problem asks you to find the minimum length of a string after repeatedly removing substrings "AB" and "CD". The input is a string `s` containing only uppercase English letters. The output is an integer representing the minimum possible length of the string after applying the removal operations. Crucially, removing a substring can create new instances of "AB" or "CD", requiring iterative removal until no more such substrings exist. The string is processed from left to right, and removals are not limited to a single pass. The constraints limit the string length to a maximum of 100 characters.

Solution Approach

The provided solutions utilize a stack (explicitly in Python and C++, implicitly in the Java solution through repeated string replacement) to efficiently track and remove "AB" and "CD" substrings. The algorithm iterates through the string. When a character is encountered that forms "AB" or "CD" with the top of the stack, both characters are removed (popped from the stack). Otherwise, the character is pushed onto the stack. Finally, the length of the remaining stack represents the minimum length of the string.

Step-by-Step Algorithm

Step 1: Initialize an empty stack (or equivalent data structure).
Step 2: Iterate through the input string, character by character.
Step 3: If the current character forms "AB" or "CD" with the top of the stack, pop the top element and the current character from the stack.
Step 4: If the current character does not form "AB" or "CD", push the character onto the stack.
Step 5: After iterating through the entire string, the length of the stack is the minimum length of the resulting string.

Key Insights

Insight 1: Using a stack (or similar data structure) to efficiently track characters and identify "AB" or "CD" subsequences is crucial for avoiding redundant checks.
Insight 2: The problem can be solved iteratively by repeatedly removing "AB" and "CD" until no more such substrings are found. A stack efficiently manages this process by only considering adjacent pairs of characters.
Insight 3: The solution needs to handle cases where removing one substring creates another. A simple one-pass check is insufficient; iterative removal (or a stack-based approach) is required.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Stack, Simulation.

Companies

Asked at: J.P. Morgan, Yelp.

Minimum String Length After Removing Substrings | LeetCode 2696

Easy

String Stack Simulation

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int minLength(String s) {
        while (s.contains("AB") || s.contains("CD")) {
            if (s.contains("AB")) {
                s = s.replaceFirst("AB", "");
            } else {
                s = s.replaceFirst("CD", "");
            }
        }
        return s.length();
    }
}

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Problem Info

Description

Problem 2696: Minimum String Length After Removing Substrings - Detailed explanation and solution approaches.

Topics

String Stack Simulation

Companies

J.P. MorganYelp

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:2696

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize an empty stack (or equivalent data structure).
2
Step 2: Iterate through the input string, character by character.
3
Step 3: If the current character forms "AB" or "CD" with the top of the stack, pop the top element and the current character from the stack.
4
Step 4: If the current character does not form "AB" or "CD", push the character onto the stack.
5
Step 5: After iterating through the entire string, the length of the stack is the minimum length of the resulting string.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The loop iterates through the string once. Stack operations (push and pop) take constant time. Therefore, the time complexity is linear. The space used is proportional to the maximum size of the stack, which is at most the length of the input string.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach

Recursively remove "AB" and "CD" until no more removals are possible. This can be less efficient than an iterative approach using a stack due to function call overhead.

Time: O(n^2) in the worst case, due to repeated string traversals. It could be optimized to O(n) with careful management of indexes.Space: O(n) in the worst case (due to recursive calls)

Trade-offs: A recursive approach might be easier to understand conceptually for beginners, but the iterative stack-based approach is more efficient.

✨

Key Insights

1Insight 1: Using a stack (or similar data structure) to efficiently track characters and identify "AB" or "CD" subsequences is crucial for avoiding redundant checks.
2Insight 2: The problem can be solved iteratively by repeatedly removing "AB" and "CD" until no more such substrings are found. A stack efficiently manages this process by only considering adjacent pairs of characters.
3Insight 3: The solution needs to handle cases where removing one substring creates another. A simple one-pass check is insufficient; iterative removal (or a stack-based approach) is required.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string (""). The solution correctly returns 0.
Edge case 2: String with only one character. The solution correctly returns 1.
Edge case 3: String containing only "AB" or "CD" repeatedly, e.g., "ABABAB". The solution correctly returns 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where removing one substring creates another. A simple one-pass check will fail for strings like "ABABCDA".

✗Common mistake 2: Inefficient string manipulation in Java (using `replaceFirst` repeatedly could lead to performance issues for long strings). Using a stack-like approach within Java would be preferable.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Text processing and data cleaning – removing unwanted patterns or substrings from text data.

🏦Real-world scenario 2: Compiler design or code optimization – eliminating redundant code sequences or simplifying expressions.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 24, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum String Length After Removing Substrings"?

For this easy-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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