String Compression - Practice Coding Problems

String Compression - Problem

String Compression Challenge - Transform Data Efficiently!

You're given a character array chars and need to compress it using a run-length encoding algorithm. The goal is to replace consecutive duplicate characters with the character followed by its count.

The Rules:
• If a character appears only once, just keep the character
• If a character repeats consecutively, write the character + count
• For counts ≥ 10, split digits into separate array positions
• Modify the input array in-place using constant extra space
• Return the new compressed length

Example: ['a','a','b','b','c','c','c'] becomes ['a','2','b','2','c','3']

This is a classic two-pointers problem that tests your ability to manipulate arrays efficiently while managing edge cases like multi-digit counts.

Input & Output

example_1.py — Basic Compression

$ Input: chars = ['a','a','b','b','c','c','c']

› Output: 6, chars = ['a','2','b','2','c','3']

💡 Note: Groups of consecutive characters: 'aa' becomes 'a2', 'bb' becomes 'b2', 'ccc' becomes 'c3'. The compressed length is 6.

example_2.py — Single Characters

$ Input: chars = ['a']

› Output: 1, chars = ['a']

💡 Note: Single character 'a' stays as 'a' since groups of length 1 don't need count notation.

example_3.py — Multi-digit Count

$ Input: chars = ['a','b','b','b','b','b','b','b','b','b','b','b','b']

› Output: 4, chars = ['a','b','1','2']

💡 Note: Single 'a' stays as 'a'. Twelve consecutive 'b's become 'b12'. Count 12 is split into digits '1' and '2'. Total compressed length is 4.

Visualization

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Understanding the Visualization

Setup Pointers

Place a 'scanner' pointer to read books and a 'organizer' pointer to write the compressed format

Count Groups

Scanner counts consecutive identical books while organizer waits

Write Compressed

Organizer writes the book type and count tag at the current position

Continue Process

Repeat until scanner reaches the end, organizer position gives final length

Key Takeaway

🎯 Key Insight: The write pointer always stays behind the read pointer, allowing safe in-place modification while maintaining the compressed format efficiently!

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array where n is the length of input array

✓ Linear Growth

Space Complexity

O(1)

Only using a constant amount of extra variables (pointers and counters)

✓ Linear Space

Constraints

1 ≤ chars.length ≤ 2000
chars[i] is a lowercase English letter, uppercase English letter, or digit
Follow-up: Could you solve it using only O(1) extra space?

Asked in

⊞ Microsoft 45 a Amazon 38 G Google 32 Ai Adobe 28

The optimal solution uses a two-pointers technique with O(1) space complexity. A read pointer scans through consecutive characters to count them, while a write pointer places the compressed result directly back into the original array. This approach handles multi-digit counts efficiently by converting numbers to strings and placing each digit separately.

Common Approaches

Approach	Time	Space	Notes
✓ Two Pointers (Optimal)	O(n)	O(1)	Use read and write pointers to compress array in-place with constant space
Brute Force (Extra String Building)	O(n)	O(n)	Build compressed string separately, then copy back to original array

Two Pointers (Optimal) — Algorithm Steps

Initialize read pointer (i) and write pointer (writeIndex) both at 0
While read pointer hasn't reached the end:
Count consecutive occurrences of current character
Write the character at writeIndex position
If count > 1, write each digit of the count
Move read pointer by count amount, advance writeIndex accordingly
Return the final writeIndex as the compressed length

Visualization

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Step-by-Step Walkthrough

Initialize Pointers

Set read pointer and write pointer both at position 0

Count Characters

Use read pointer to count consecutive identical characters

Write Compressed

Use write pointer to place character and count in compressed form

Advance Pointers

Move read pointer by count amount, advance write pointer accordingly

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

int compress(char* chars, int charsSize) {
    if (charsSize == 0) return 0;
    
    int write = 0;  // Write pointer
    int i = 0;      // Read pointer
    
    while (i < charsSize) {
        char currentChar = chars[i];
        int count = 1;
        
        // Count consecutive characters
        while (i + count < charsSize && chars[i + count] == currentChar) {
            count++;
        }
        
        // Write the character
        chars[write] = currentChar;
        write++;
        
        // Write count if greater than 1
        if (count > 1) {
            // Convert count to string and write each digit
            char countStr[12];
            sprintf(countStr, "%d", count);
            for (int j = 0; j < strlen(countStr); j++) {
                chars[write] = countStr[j];
                write++;
            }
        }
        
        // Move read pointer
        i += count;
    }
    
    return write;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array where n is the length of input array

✓ Linear Growth

Space Complexity

O(1)

Only using a constant amount of extra variables (pointers and counters)

✓ Linear Space

Constraints

1 ≤ chars.length ≤ 2000
chars[i] is a lowercase English letter, uppercase English letter, or digit
Follow-up: Could you solve it using only O(1) extra space?

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Two Pointers (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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