What is the time complexity of Remove All Adjacent Duplicates in String II?

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

What data structures are used to solve Remove All Adjacent Duplicates in String II?

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

What are the key insights for solving Remove All Adjacent Duplicates in String II?

Insight 1: Using a stack is an efficient way to keep track of characters and their counts, allowing for easy removal of duplicates when the count reaches `k`. Insight 2: The stack stores character-count pairs. When a new character is encountered, either the count of the last character in the stack increases (if it's the same character), or a new character-count pair is added to the stack.

Which companies ask LeetCode 1209. Remove All Adjacent Duplicates in String II in interviews?

This problem is frequently asked at Attentive, Disney, FactSet, Grammarly.

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Remove All Adjacent Duplicates in String II - LeetCode 1209 Solution

Remove All Adjacent Duplicates in String II - Complete Solution Guide

Remove All Adjacent Duplicates in String II is LeetCode problem 1209, a Medium 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 and an integer k , a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them, causing the left and the right side of the deleted substring to concatenate together. We repeatedly make k duplicate removals on s until we no longer can. Return the final string after all such duplicate removals have been made . It is guaranteed that the answer is unique . Example 1: Input: s = "abcd", k = 2 Output: "abcd" Explanation: There's nothing to

Detailed Explanation

The problem requires you to repeatedly remove all adjacent duplicate characters of length `k` from a given string `s` until no more removals are possible. The goal is to return the final string after all such removals. For instance, if `s = "deeedbbcccbdaa"` and `k = 3`, you would first remove "eee" and "ccc" resulting in "ddbbbdaa", then remove "bbb" to get "dddaa", and finally remove "ddd" which results in the final string "aa".

Solution Approach

The solution uses a stack to keep track of the characters and their respective counts. It iterates through the input string `s`. For each character, it checks if the stack is empty or if the current character is different from the character at the top of the stack. If either is true, it adds a new pair (character, 1) to the stack. If the current character is the same as the character at the top of the stack, it increments the count of that character. After each character is processed, it checks if the count of the character at the top of the stack has reached `k`. If it has, the character-count pair is removed from the stack. Finally, the characters remaining in the stack are concatenated to form the final string.

Step-by-Step Algorithm

Step 1: Initialize an empty stack to store character-count pairs.
Step 2: Iterate through the input string `s` character by character.
Step 3: For each character, check if the stack is empty or if the current character is different from the character at the top of the stack.
Step 4: If the stack is empty or the characters are different, push a new character-count pair (character, 1) onto the stack.
Step 5: If the characters are the same, increment the count of the character at the top of the stack.
Step 6: After each character is processed, check if the count of the character at the top of the stack is equal to `k`.
Step 7: If the count is equal to `k`, pop the character-count pair from the stack.
Step 8: After processing all characters in the input string, construct the final string by concatenating the remaining characters in the stack based on their respective counts.
Step 9: Return the final string.

Key Insights

Insight 1: Using a stack is an efficient way to keep track of characters and their counts, allowing for easy removal of duplicates when the count reaches `k`.
Insight 2: The stack stores character-count pairs. When a new character is encountered, either the count of the last character in the stack increases (if it's the same character), or a new character-count pair is added to the stack.
Insight 3: Maintaining the character-count pairs lets us quickly identify and remove duplicates of length 'k' without having to re-scan the string from the beginning.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Stack.

Companies

Asked at: Attentive, Disney, FactSet, Grammarly.

Remove All Adjacent Duplicates in String II | LeetCode 1209

Medium

String Stack

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public String removeDuplicates(String s, int k) {
        Stack<int[]> stack = new Stack<>(); // Stores [character, count] pairs

        for (char char_ : s.toCharArray()) {
            if (!stack.isEmpty() && stack.peek()[0] == char_) {
                stack.peek()[1] += 1;
            } else {
                stack.push(new int[]{char_, 1});
            }
            
            if (!stack.isEmpty() && stack.peek()[1] == k) {
                stack.pop();
            }
        }
        
        StringBuilder result = new StringBuilder();
        for (int[] pair : stack) {
            char c = (char) pair[0];
            int count = pair[1];
            for (int i = 0; i < count; i++) {
                result.append(c);
            }
        }
        
        return result.toString();
    }
}

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Problem Info

Description

Problem 1209: Remove All Adjacent Duplicates in String II - Detailed explanation and solution approaches.

Topics

String Stack

Companies

AttentiveDisneyFactSetGrammarly

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:1209

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

9 steps

1
Step 1: Initialize an empty stack to store character-count pairs.
2
Step 2: Iterate through the input string `s` character by character.
3
Step 3: For each character, check if the stack is empty or if the current character is different from the character at the top of the stack.
4
Step 4: If the stack is empty or the characters are different, push a new character-count pair (character, 1) onto the stack.
5
Step 5: If the characters are the same, increment the count of the character at the top of the stack.
6
Step 6: After each character is processed, check if the count of the character at the top of the stack is equal to `k`.
7
Step 7: If the count is equal to `k`, pop the character-count pair from the stack.
8
Step 8: After processing all characters in the input string, construct the final string by concatenating the remaining characters in the stack based on their respective counts.
9
Step 9: Return the final 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 time complexity is linear because each character in the string is processed at most a constant number of times (pushed onto and potentially popped from the stack). The space complexity is also linear because, in the worst case, the stack could store a character and count for each character in 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

Two Pointers

An alternative approach could use two pointers, a read pointer and a write pointer, to simulate the stack behavior in place. This would reduce the space complexity.

Time: O(n)Space: O(n) in the worst-case to store counts, and O(1) if the count can be combined with the string itself (in-place replacement)

Trade-offs: The two-pointer approach might be more complex to implement correctly. It may need extra space to store character counts or perform in-place operations. Using a stack offers better readability and maintainability.

✨

Key Insights

1Insight 1: Using a stack is an efficient way to keep track of characters and their counts, allowing for easy removal of duplicates when the count reaches `k`.
2Insight 2: The stack stores character-count pairs. When a new character is encountered, either the count of the last character in the stack increases (if it's the same character), or a new character-count pair is added to the stack.
3Insight 3: Maintaining the character-count pairs lets us quickly identify and remove duplicates of length 'k' without having to re-scan the string from the beginning.

📚

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 input string. The algorithm correctly handles an empty string because the loop won't execute, and an empty string will be returned.
Edge case 2: `k` is greater than the length of the string. The algorithm correctly processes such cases. If no characters are duplicated at least `k` times, the original string will be returned.
Edge case 3: The entire string can be removed. For example, s = "aaaa", k = 2, the result will be "". The algorithm correctly handles this scenario.
Edge case 4: String with no duplicates. If the string contains no duplicates, the stack will eventually contain all characters. The algorithm correctly returns such strings. For example, s = "abcd", k = 2, the result is "abcd".

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to pop the element from the stack when its count reaches `k`. This will result in an incorrect output string.

✗Common mistake 2: Not handling the case where the stack is empty when trying to access the top element, resulting in errors.

✗Common mistake 3: Prematurely optimizing before ensuring the core logic is correct can lead to harder-to-debug code. Focus on correctness first.

✅

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: Data compression. Similar duplicate removal techniques are used in lossless data compression algorithms to reduce redundancy and file sizes.

🏦Real-world scenario 2: Text processing. Removing consecutive duplicate characters or words can be useful in cleaning and normalizing text data for natural language processing tasks.

🎮Real-world scenario 3: Game development. Removing redundant actions or events in a game's event stream can optimize performance and reduce storage requirements.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: October 8, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Remove All Adjacent Duplicates in String II"?

For this medium-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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