What is the time complexity of Lexicographically Minimum String After Removing Stars?

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

What data structures are used to solve Lexicographically Minimum String After Removing Stars?

This problem uses Hash Table, String, Stack, Greedy, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Lexicographically Minimum String After Removing Stars?

Insight 1: The order of processing '*' characters matters since removing characters earlier affects what's available for later removals. Insight 2: We can greedily remove the leftmost smallest available character for each '*'. This is key for obtaining the lexicographically smallest final string.

Which companies ask LeetCode 3170. Lexicographically Minimum String After Removing Stars in interviews?

This problem is frequently asked at Flexera.

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Lexicographically Minimum String After Removing Stars - LeetCode 3170 Solution

Q: How to solve Lexicographically Minimum String After Removing Stars?

The solution uses a greedy approach. It iterates through the string, and for each character: If it's a '*', it searches for the leftmost smallest non-'*' character that hasn't been removed yet and removes it, along with the '*'. If it's a regular character, it stores its index for later removal if a '*' is encountered. A boolean array keeps track of removed characters to avoid repeated string modifications. The indexes of each character is saved in a list of stacks, where index of stack represents the ascii index of the char.

Lexicographically Minimum String After Removing Stars - Complete Solution Guide

Lexicographically Minimum String After Removing Stars is LeetCode problem 3170, 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 . It may contain any number of '*' characters. Your task is to remove all '*' characters. While there is a '*' , do the following operation: Delete the leftmost '*' and the smallest non- '*' character to its left . If there are several smallest characters, you can delete any of them. Return the lexicographically smallest resulting string after removing all '*' characters. Example 1: Input: s = "aaba*" Output: "aab" Explanation: We should delete one of the 'a' characters

Detailed Explanation

The problem requires you to process a string `s` containing lowercase English letters and asterisks ('*'). The goal is to remove all '*' characters following a specific rule: for each '*' encountered, you must also remove the leftmost smallest non-'*' character to its left. The 'smallest' character refers to its lexicographical order (a < b < c, etc.). The final result should be the lexicographically smallest string after all '*' characters have been processed and removed.

Solution Approach

The solution uses a greedy approach. It iterates through the string, and for each character: If it's a '*', it searches for the leftmost smallest non-'*' character that hasn't been removed yet and removes it, along with the '*'. If it's a regular character, it stores its index for later removal if a '*' is encountered. A boolean array keeps track of removed characters to avoid repeated string modifications. The indexes of each character is saved in a list of stacks, where index of stack represents the ascii index of the char.

Step-by-Step Algorithm

Step 1: Initialize a list of stacks, 'positions', where each stack at index 'i' stores the indices of the character 'a' + i in the original string. Also, initialize a boolean array 'is_removed' to track characters that have been removed.
Step 2: Iterate through the string 's'.
Step 3: If the current character is '*', mark it as removed (is_removed[i] = true). Iterate through the positions list in the range 0-25. If stack at index j is not empty, pop the top element from this stack (which will be the last occurence of this character to the left of '*'). Mark this characters position as removed using is_removed[idx_to_remove] = true, and break out of the inner loop.
Step 4: If the current character is not a '*', calculate its index (char - 'a') and push the current index i to the corresponding stack in the positions array.
Step 5: Create the final result string by iterating through the original string and appending characters that are not marked as removed in the is_removed array.
Step 6: Return the resulting string.

Key Insights

Insight 1: The order of processing '*' characters matters since removing characters earlier affects what's available for later removals.
Insight 2: We can greedily remove the leftmost smallest available character for each '*'. This is key for obtaining the lexicographically smallest final string.
Insight 3: Maintaining the positions of characters and a separate flag array for removed characters avoids repeated string manipulations, which are costly.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Stack, Greedy, Heap (Priority Queue).

Companies

Asked at: Flexera.

Lexicographically Minimum String After Removing Stars | LeetCode 3170

Medium

Hash Table String Stack Greedy Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

class Solution {
    public String clearStars(String s) {
        int n = s.length();
        List<Stack<Integer>> positions = new ArrayList<>();
        for (int i = 0; i < 26; i++) {
            positions.add(new Stack<>());
        }
        boolean[] is_removed = new boolean[n];

        for (int i = 0; i < n; i++) {
            char charAtI = s.charAt(i);
            if (charAtI == '*') {
                is_removed[i] = true;
                for (int j = 0; j < 26; j++) {
                    if (!positions.get(j).isEmpty()) {
                        int idx_to_remove = positions.get(j).pop();
                        is_removed[idx_to_remove] = true;
                        break;
                    }
                }
            } else {
                positions.get(charAtI - 'a').push(i);
            }
        }

        StringBuilder result = new StringBuilder();
        for (int i = 0; i < n; i++) {
            if (!is_removed[i]) {
                result.append(s.charAt(i));
            }
        }

        return result.toString();
    }
}

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Problem Info

Description

Problem 3170: Lexicographically Minimum String After Removing Stars - Detailed explanation and solution approaches.

Topics

Hash Table String Stack Greedy Heap (Priority Queue)

Companies

Flexera

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3170

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 a list of stacks, 'positions', where each stack at index 'i' stores the indices of the character 'a' + i in the original string. Also, initialize a boolean array 'is_removed' to track characters that have been removed.
2
Step 2: Iterate through the string 's'.
3
Step 3: If the current character is '*', mark it as removed (is_removed[i] = true). Iterate through the positions list in the range 0-25. If stack at index j is not empty, pop the top element from this stack (which will be the last occurence of this character to the left of '*'). Mark this characters position as removed using is_removed[idx_to_remove] = true, and break out of the inner loop.
4
Step 4: If the current character is not a '*', calculate its index (char - 'a') and push the current index i to the corresponding stack in the positions array.
5
Step 5: Create the final result string by iterating through the original string and appending characters that are not marked as removed in the is_removed array.
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 derived from the single pass through the string and the inner loop of constant size. The space complexity comes from storing, in the worst case, the indices of all characters and the boolean removal flags. The final output will be max n in size too.

📖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

Stack-based Approach

Use a stack to keep track of characters. When a '*' is encountered, pop the smallest available character from the stack. This involves sorting the stack or using a priority queue to ensure removal of the smallest character.

Time: O(n log k) where k is the max number of chars to the left of '*'Space: O(n)

Trade-offs: Stack with sorting provides a cleaner, but less efficient approach. Using a priority queue within the stack solution might improve runtime to O(n log alphabet_size) which is O(n) in practice but has worse constant factors.

✨

Key Insights

1Insight 1: The order of processing '*' characters matters since removing characters earlier affects what's available for later removals.
2Insight 2: We can greedily remove the leftmost smallest available character for each '*'. This is key for obtaining the lexicographically smallest final string.
3Insight 3: Maintaining the positions of characters and a separate flag array for removed characters avoids repeated string manipulations, which are costly.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Stack

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

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

Edge Cases

Edge case 1: Empty string. The code implicitly handles this as the loops won't execute, and an empty string will be returned.
Edge case 2: String with no '*' characters. The 'is_removed' array will remain all false, and the original string will be returned, which is the correct behavior.
Edge case 3: String with all '*' characters. The code will attempt to remove characters to the left, which may not exist, this is handled as each stack in positions can be empty.
Edge case 4: String with consecutive '*' characters. Each star will remove available char from left greedily, satisfying problem constraints.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the lexicographical order when choosing which character to remove. This will result in a non-lexicographically smallest string.

✗Common mistake 2: Modifying the string directly within the loop can lead to incorrect indexing and inefficient string operations. Using a boolean array to track removed characters is more efficient.

✗Common mistake 3: Not handling the case where there might not be a character to the left of a '*' to remove. The code should gracefully handle empty stacks.

✗Common mistake 4: Overlooking the 'leftmost' requirement. If there are multiple occurrences of the smallest character, you must remove the one closest to the beginning of the 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 where users perform undo operations that involve deleting characters based on context. The '*' can represent a trigger to selectively undo based on other conditions.

🏦Real-world scenario 2: Data cleaning and preprocessing where specific characters need to be removed based on the presence of special markers. This could be applied to log file processing or sanitizing user input.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

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

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

👨‍💻 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: September 19, 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

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 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 "Lexicographically Minimum String After Removing Stars"?

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

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