What is the time complexity of Removing Stars From a String?

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

What data structures are used to solve Removing Stars From a String?

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

What are the key insights for solving Removing Stars From a String?

Insight 1: The problem can be efficiently solved by iterating through the string and maintaining a stack (or a dynamically resizing array) to simulate the removal process. Insight 2: The problem requires processing the string sequentially, so the order of characters matters significantly.

Which companies ask LeetCode 2390. Removing Stars From a String in interviews?

This problem is frequently asked at IBM.

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Removing Stars From a String - LeetCode 2390 Solution

Removing Stars From a String - Complete Solution Guide

Removing Stars From a String is LeetCode problem 2390, 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 , which contains stars * . In one operation, you can: Choose a star in s . Remove the closest non-star character to its left , as well as remove the star itself. Return the string after all stars have been removed . Note: The input will be generated such that the operation is always possible. It can be shown that the resulting string will always be unique. Example 1: Input: s = "leet**cod*e" Output: "lecoe" Explanation: Performing the removals from left to right: - The c

Detailed Explanation

The problem asks us to remove all stars from a given string 's'. Each star signifies a removal operation. When a star is encountered, we must remove the closest non-star character immediately to its left, as well as the star itself. The task is to return the modified string after all such removals are completed.

Solution Approach

The solution uses a stack-like data structure (implemented as a list/vector/stack in different languages) to keep track of characters that are not yet removed. When a character is encountered, it is either added to the stack if it's not a star, or the top element of the stack is removed if it's a star (provided the stack is not empty). Finally, the remaining characters in the stack form the modified string.

Step-by-Step Algorithm

Step 1: Initialize an empty list (or stack, vector, depending on the language) called 'res' to store the result.
Step 2: Iterate through the input string 's' character by character.
Step 3: If the current character is a star ('*'), check if the 'res' list is not empty. If it's not empty, remove the last element from the 'res' list (simulating the removal of the closest non-star character to the left of the star).
Step 4: If the current character is not a star, append it to the 'res' list.
Step 5: After iterating through the entire string, join the characters in the 'res' list to form the final modified string.
Step 6: Return the resulting string.

Key Insights

Insight 1: The problem can be efficiently solved by iterating through the string and maintaining a stack (or a dynamically resizing array) to simulate the removal process.
Insight 2: The problem requires processing the string sequentially, so the order of characters matters significantly.
Insight 3: The constraints guarantee that the operation is always possible, meaning there's always a non-star character to the left of any star that needs removing.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Stack, Simulation.

Companies

Asked at: IBM.

Removing Stars From a String | LeetCode 2390

Medium

String Stack Simulation

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public String removeStars(String s) {
        Stack<Character> res = new Stack<>();
        for (char char_ : s.toCharArray()) {
            if (char_ == '*') {
                if (!res.isEmpty()) {
                    res.pop();
                }
            } else {
                res.push(char_);
            }
        }
        StringBuilder sb = new StringBuilder();
        while (!res.isEmpty()) {
            sb.insert(0, res.pop());
        }
        return sb.toString();
    }
}

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Problem Info

Description

Problem 2390: Removing Stars From a String - Detailed explanation and solution approaches.

Topics

String Stack Simulation

Companies

IBM

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2390

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

6 steps

1
Step 1: Initialize an empty list (or stack, vector, depending on the language) called 'res' to store the result.
2
Step 2: Iterate through the input string 's' character by character.
3
Step 3: If the current character is a star ('*'), check if the 'res' list is not empty. If it's not empty, remove the last element from the 'res' list (simulating the removal of the closest non-star character to the left of the star).
4
Step 4: If the current character is not a star, append it to the 'res' list.
5
Step 5: After iterating through the entire string, join the characters in the 'res' list to form the final modified string.
6
Step 6: Return 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 time complexity is determined by the single pass through the string. The space complexity depends on the number of non-star characters that are stored in the 'res' list. In the worst-case scenario, the list 'res' will contain almost all of the characters from the input string, leading to O(n) space usage.

📖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 (In-place, if allowed to modify input)

Use two pointers: one to iterate through the string and another to keep track of the index of the next character to be placed in the modified string. When encountering a star, decrement the second pointer. Otherwise, copy the character to the second pointer's location. Requires being able to modify the original string.

Time: O(n)Space: O(1) if modification of the input string is allowed, O(n) otherwise (to create a copy and avoid modifying the input).

Trade-offs: More space efficient than using a stack if modifying the input string is allowed, but modifies the original string, which might not be desirable. More complicated logic with pointers compared to using stack.

✨

Key Insights

1Insight 1: The problem can be efficiently solved by iterating through the string and maintaining a stack (or a dynamically resizing array) to simulate the removal process.
2Insight 2: The problem requires processing the string sequentially, so the order of characters matters significantly.
3Insight 3: The constraints guarantee that the operation is always possible, meaning there's always a non-star character to the left of any star that needs removing.

📚

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 solution will correctly return an empty string since the loop won't execute.
Edge case 2: Input string with only stars. The list/stack will remain empty throughout the process, resulting in an empty string as output. This is because for each star, we attempt to pop from an empty stack.
Edge case 3: Input string starting with stars. The solution handles it by checking if the stack is empty before popping, preventing any errors.
Edge case 4: Input string without stars. The result will be the same as the input. All characters are added to the stack.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the stack is empty when encountering a star. This will lead to an error (e.g., popping from an empty stack). The code should always check if the stack is empty before attempting to pop an element.

✗Common mistake 2: Incorrectly joining the characters in the 'res' list/vector. Some languages require special methods to convert a list of characters back into a string.

✗Common mistake 3: Failing to allocate memory correctly and handle strings correctly when using C. Using `strcpy` requires careful memory management.

✗Common mistake 4: Getting confused about if you need a stack versus directly building a result string. Often, you only need to directly build result string.

✅

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 editors or IDEs often have features like undo/redo. The stack-based approach is conceptually similar to how these features are implemented, where each action is pushed onto a stack, and undoing involves popping actions from the stack.

🏦Real-world scenario 2: Parsing expressions or code where certain symbols need to be matched and removed (e.g., matching parentheses). Although this problem is simpler, the stack principle applies.

🎮Real-world scenario 3: Backspace functionality. Every time a backspace key is pressed, the previously entered character is removed, analogous to the star removal operation. Stack is a natural fit.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: July 21, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

❓

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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Removing Stars From a String"?

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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