What is the time complexity of String Compression III?

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

What data structures are used to solve String Compression III?

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

What are the key insights for solving String Compression III?

Insight 1: The core idea is to iterate through the input string and identify consecutive segments of the same character. Insight 2: We need to limit the length of each segment to a maximum of 9, as only single-digit counts are allowed in the compressed string.

Which companies ask LeetCode 3163. String Compression III in interviews?

This problem is frequently asked at Affirm, Qualcomm.

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String Compression III - LeetCode 3163 Solution

String Compression III - Complete Solution Guide

String Compression III is LeetCode problem 3163, 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

Given a string word , compress it using the following algorithm: Begin with an empty string comp . While word is not empty, use the following operation: Remove a maximum length prefix of word made of a single character c repeating at most 9 times. Append the length of the prefix followed by c to comp . Return the string comp . Example 1: Input: word = "abcde" Output: "1a1b1c1d1e" Explanation: Initially, comp = "" . Apply the operation 5 times, choosing "a" , "b" , "c" , "d" , and "e" as the pref

Detailed Explanation

The problem requires us to compress a given string 'word' by replacing consecutive repeating characters with the count of their repetitions followed by the character itself. The repetitions must be handled in chunks of at most 9 consecutive characters. The compressed string 'comp' is built iteratively from left to right by identifying the maximum-length prefix composed of a single character repeated at most 9 times, appending the length of the prefix and the character to 'comp', and repeating the process until 'word' is empty. For example, 'aaaaaaaaaaaaaabb' becomes '9a5a2b' because there are twelve 'a's initially which are split into a '9a' and a '5a', followed by '2b'.

Solution Approach

The provided solution uses a simple iterative approach. It iterates through the input string, tracking the current character and its count. When the character changes or the count reaches 9, it appends the count and the character to the result string. The process repeats until the entire input string has been processed.

Step-by-Step Algorithm

Step 1: Initialize an empty result string (or a list of characters in Python for better performance during appending).
Step 2: Initialize an index 'i' to 0 to traverse the input string.
Step 3: While 'i' is less than the length of the input string, repeat steps 4-7.
Step 4: Identify the character at the current index 'i'.
Step 5: Initialize a counter 'count' to 0 and another index 'j' to 'i'.
Step 6: Increment 'j' as long as 'j' is within the bounds of the string, the character at 'j' is the same as the identified character, and the count is less than 9.
Step 7: Append the 'count' (converted to string) and the identified character to the result string.
Step 8: Update 'i' to the current value of 'j', effectively skipping the processed prefix.
Step 9: Once the loop finishes, convert the result list to a string (if using Python list) and return the string.

Key Insights

Insight 1: The core idea is to iterate through the input string and identify consecutive segments of the same character.
Insight 2: We need to limit the length of each segment to a maximum of 9, as only single-digit counts are allowed in the compressed string.
Insight 3: Efficient string manipulation and concatenation are crucial for optimal performance.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String.

Companies

Asked at: Affirm, Qualcomm.

String Compression III | LeetCode 3163

Medium

String

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

Solution Code

Available in:

class Solution {
    public String compressedString(String word) {
        StringBuilder res = new StringBuilder();
        int i = 0;
        int n = word.length();
        
        while (i < n) {
            char char_ = word.charAt(i);
            int count = 0;
            
            int j = i;
            while (j < n && word.charAt(j) == char_ && count < 9) {
                count += 1;
                j += 1;
            }
            
            res.append(String.valueOf(count));
            res.append(char_);
            
            i = j;
            
        }
        
        return res.toString();
    }
}

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Problem Info

Description

Problem 3163: String Compression III - Detailed explanation and solution approaches.

Topics

String

Companies

AffirmQualcomm

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:3163

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 result string (or a list of characters in Python for better performance during appending).
2
Step 2: Initialize an index 'i' to 0 to traverse the input string.
3
Step 3: While 'i' is less than the length of the input string, repeat steps 4-7.
4
Step 4: Identify the character at the current index 'i'.
5
Step 5: Initialize a counter 'count' to 0 and another index 'j' to 'i'.
6
Step 6: Increment 'j' as long as 'j' is within the bounds of the string, the character at 'j' is the same as the identified character, and the count is less than 9.
7
Step 7: Append the 'count' (converted to string) and the identified character to the result string.
8
Step 8: Update 'i' to the current value of 'j', effectively skipping the processed prefix.
9
Step 9: Once the loop finishes, convert the result list to a string (if using Python list) and return the 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 algorithm iterates through the string once. The inner 'while' loop does not add to the time complexity because in total, each character of input string is visited once. The compressed string can at most be twice as long as the original string (e.g., 'abc' becomes '1a1b1c'). Therefore, the space complexity is linear.

📖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

Regular Expressions

Use regular expressions to find consecutive repeating characters and their counts. This could be more concise but potentially less efficient depending on the regex engine.

Time: O(n) in most regex implementationsSpace: O(n) for the result string

Trade-offs: Might be more readable for those familiar with regex, but could be slower due to regex overhead. Less control over limiting repetition to 9.

✨

Key Insights

1Insight 1: The core idea is to iterate through the input string and identify consecutive segments of the same character.
2Insight 2: We need to limit the length of each segment to a maximum of 9, as only single-digit counts are allowed in the compressed string.
3Insight 3: Efficient string manipulation and concatenation are crucial for optimal performance.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Empty string - The code handles this gracefully by returning an empty string because the while loop condition (i < n) will be immediately false.
Edge case 2: String with a single character repeated more than 9 times (e.g., 'aaaaaaaaaa') - The code correctly splits this into chunks of 9 and the remaining characters, resulting in '9a1a'.
Edge case 3: String with alternating characters (e.g., 'abcabcabc') - The code correctly compresses this into '1a1b1c1a1b1c1a1b1c'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to limit the repetition count to 9. This would lead to incorrect compressed strings if a character repeats more than 9 times consecutively.

✗Common mistake 2: Incorrectly updating the index 'i' after processing a prefix. This could lead to skipping characters or infinite loops.

✗Common mistake 3: Not handling empty input string properly. The code should return an empty string in this case.

✅

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 algorithms. This problem embodies a simplified form of run-length encoding (RLE), which is used in image and video compression formats.

🏦Real-world scenario 2: Log file analysis. When analyzing log files, repeated patterns (e.g., timestamp patterns, error messages) can be compressed for easier storage and analysis, and can highlight recurring issues

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: September 12, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

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

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

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.

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 "String Compression III"?

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 standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?