What is the time complexity of Apply Operations to Make String Empty?

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

What data structures are used to solve Apply Operations to Make String Empty?

This problem uses Array, Hash Table, Sorting, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Apply Operations to Make String Empty?

Insight 1: The key insight is recognizing that we only need to find the characters with the maximum frequency in the original string and maintain their original order (from right to left). This is because these characters will survive the most rounds of removal, making up the final non-empty string. Insight 2: Instead of simulating the entire process, which would be inefficient, we can use a frequency map (or counter) to determine which characters have the highest frequency.

Which companies ask LeetCode 3039. Apply Operations to Make String Empty in interviews?

This problem is frequently asked at Virtusa.

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Apply Operations to Make String Empty - LeetCode 3039 Solution

Apply Operations to Make String Empty - Complete Solution Guide

Apply Operations to Make String Empty is LeetCode problem 3039, 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 . Consider performing the following operation until s becomes empty : For every alphabet character from 'a' to 'z' , remove the first occurrence of that character in s (if it exists). For example, let initially s = "aabcbbca" . We do the following operations: Remove the underlined characters s = " a a bc bbca" . The resulting string is s = "abbca" . Remove the underlined characters s = " ab b c a" . The resulting string is s = "ba" . Remove the underlined characters s =

Detailed Explanation

The problem asks us to simulate a series of operations on a given string `s`. In each operation, we iterate through the lowercase English alphabet ('a' to 'z') and remove the first occurrence of each letter present in `s`. We repeat this operation until the string `s` is empty. The goal is to return the string `s` right *before* the last operation is applied (i.e., the string that is about to become empty). The string contains only lowercase characters and its length can be up to 5 * 10^5.

Solution Approach

The solution efficiently determines the characters present in the string just before the string becomes empty. It avoids simulation by first computing the frequency of each character in the string. Then, it identifies the maximum frequency. Finally, it iterates the string from right to left, selecting characters that have the maximum frequency, and storing them in the order they are encountered from right to left. The final result is then reversed to maintain the original order of the selected characters in the string before the last removal operation.

Step-by-Step Algorithm

Step 1: Create a frequency map (or counter) to store the count of each character in the string `s`.
Step 2: Find the maximum frequency among all character counts.
Step 3: Iterate through the string `s` from right to left.
Step 4: For each character, check if its frequency equals the maximum frequency and if it has not already been added to the result.
Step 5: If both conditions are met, add the character to a list (or array) and mark it as found.
Step 6: Stop iterating when all characters with maximum frequency have been found.
Step 7: Reverse the list of characters to restore the original order, and return the resulting string.

Key Insights

Insight 1: The key insight is recognizing that we only need to find the characters with the maximum frequency in the original string and maintain their original order (from right to left). This is because these characters will survive the most rounds of removal, making up the final non-empty string.
Insight 2: Instead of simulating the entire process, which would be inefficient, we can use a frequency map (or counter) to determine which characters have the highest frequency.
Insight 3: Since the operations are repeated until the string is empty, the last string remaining will be formed only from the characters that appear the maximum number of times.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Sorting, Counting.

Companies

Asked at: Virtusa.

Apply Operations to Make String Empty | LeetCode 3039

Medium

Array Hash Table Sorting Counting

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public String lastNonEmptyString(String s) {
        Map<Character, Integer> counts = new HashMap<>();
        for (char c : s.toCharArray()) {
            counts.put(c, counts.getOrDefault(c, 0) + 1);
        }

        int max_freq = 0;
        if (s.isEmpty()) {
            return "";
        }
        for (int freq : counts.values()) {
            max_freq = Math.max(max_freq, freq);
        }

        int final_chars_count = 0;
        for (int freq : counts.values()) {
            if (freq == max_freq) {
                final_chars_count += 1;
            }
        }

        List<Character> result_chars = new ArrayList<>();
        Set<Character> found_chars = new HashSet<>();

        for (int i = s.length() - 1; i >= 0; i--) {
            char charAtI = s.charAt(i);
            if (counts.get(charAtI) == max_freq && !found_chars.contains(charAtI)) {
                result_chars.add(charAtI);
                found_chars.add(charAtI);
                if (found_chars.size() == final_chars_count) {
                    break;
                }
            }
        }

        Collections.reverse(result_chars);
        StringBuilder sb = new StringBuilder();
        for(char c : result_chars){
            sb.append(c);
        }

        return sb.toString();
    }
}

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Problem Info

Description

Problem 3039: Apply Operations to Make String Empty - Detailed explanation and solution approaches.

Topics

Array Hash Table Sorting Counting

Companies

Virtusa

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3039

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

7 steps

1
Step 1: Create a frequency map (or counter) to store the count of each character in the string `s`.
2
Step 2: Find the maximum frequency among all character counts.
3
Step 3: Iterate through the string `s` from right to left.
4
Step 4: For each character, check if its frequency equals the maximum frequency and if it has not already been added to the result.
5
Step 5: If both conditions are met, add the character to a list (or array) and mark it as found.
6
Step 6: Stop iterating when all characters with maximum frequency have been found.
7
Step 7: Reverse the list of characters to restore the original order, and 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 dominated by the iterations through the string to build the frequency map and find the characters with maximum frequency. The space complexity is relatively small because the number of unique characters (and thus the size of the frequency map) is limited to the size of the alphabet.

📖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

Simulating the operations

Simulate the operations by repeatedly iterating through the alphabet and removing the first occurrence of each character. This is very slow because it would result in repeated linear scans and removals.

Time: O(n*k), where n is the initial length of the string and k is the number of operations required to empty the string (at most the original length of s).Space: O(n), where n is the length of the original string (in the worst case).

Trade-offs: This approach is straightforward to understand but has a significantly higher time complexity, making it unsuitable for large input strings. It is preferable only when the input size is very small or when simplicity is prioritized over performance.

✨

Key Insights

1Insight 1: The key insight is recognizing that we only need to find the characters with the maximum frequency in the original string and maintain their original order (from right to left). This is because these characters will survive the most rounds of removal, making up the final non-empty string.
2Insight 2: Instead of simulating the entire process, which would be inefficient, we can use a frequency map (or counter) to determine which characters have the highest frequency.
3Insight 3: Since the operations are repeated until the string is empty, the last string remaining will be formed only from the characters that appear the maximum number of times.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty string. The solution correctly returns an empty string when the input string `s` is empty.
Edge case 2: String with only one unique character. The solution correctly identifies the character and includes it in the result.
Edge case 3: String with all unique characters. The solution correctly identifies the characters and returns the string in original order.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating or handling the frequency of characters, which will result in a wrong final string.

✗Common mistake 2: Not considering the edge case of an empty string, leading to potential errors.

✗Common mistake 3: Not preserving the original order of characters with maximum frequency, which is crucial for returning the correct result. Reversing the list of `result_chars` is essential in this problem.

✗Common mistake 4: Simulating the operations for too many rounds when the string has been already emptied. This wastes computing resources and slows down the entire execution.

✅

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. This algorithm can be adapted to find the most frequent characters in a text, which is useful in Huffman coding and other data compression techniques.

🏦Real-world scenario 2: Analyzing log files. Identify the most frequent error codes or messages in log files to pinpoint potential issues or bottlenecks in a system.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

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

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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 7, 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 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Apply Operations to Make String Empty"?

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

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