Run-Length Decoder - Problem

Decode a run-length encoded string where consecutive identical characters are represented as count+character.

For example: 3a2b5c becomes aaabbccccc

The input string contains digits followed by letters. Multi-digit counts are supported (e.g., 12a means 12 'a' characters).

Handle invalid input gracefully:

Empty or null strings should return empty string
Invalid format (letter before digit, etc.) should return empty string
Zero counts (e.g., 0a) should be ignored

Input & Output

Example 1 — Basic Case

$ Input: encoded = "3a2b5c"

› Output: "aaabbccccc"

💡 Note: 3 'a's + 2 'b's + 5 'c's = 'aaabbccccc'

Example 2 — Multi-digit Count

$ Input: encoded = "12a3b"

› Output: "aaaaaaaaaaabbbb"

💡 Note: 12 'a's followed by 3 'b's

Example 3 — Invalid Format

$ Input: encoded = "3ab2"

› Output: ""

💡 Note: Invalid: number not followed by single character

Constraints

0 ≤ encoded.length ≤ 10⁴
encoded contains only digits and English letters
Count values can be multi-digit (1-100)
Zero counts should be ignored

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to parse the string in a structured way, clearly separating number parsing from character expansion. The optimal approach uses a state machine to process the string in one pass. Time: O(n + m), Space: O(m) where n is input length and m is output length.

Common Approaches

✓ Character-by-Character Parsing

⏱️ Time: O(n + m) Space: O(m)

Iterate through each character, trying to distinguish between digits and letters on the fly. Build number strings character by character and handle expansion when encountering letters.

Single Pass State Machine

⏱️ Time: O(n + m) Space: O(m)

Use a state machine approach to cleanly separate number parsing from character expansion. Track current state and handle transitions efficiently.

Character-by-Character Parsing — Algorithm Steps

Step 1: Loop through each character in the string
Step 2: If digit, append to current number string
Step 3: If letter, convert number string to integer and repeat the letter
Step 4: Reset number string and continue

Visualization

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Step-by-Step Walkthrough

Scan Character

Read each character and determine if digit or letter

Build Number

Accumulate digits to form complete count

Expand

When letter found, repeat it count times

Code -

solution.c — C

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

char* solution(char* encoded) {
    if (!encoded || strlen(encoded) == 0) {
        char* result = malloc(1);
        result[0] = '\0';
        return result;
    }
    
    char* result = malloc(10000);  // Assume max result length
    char currentNum[20] = {0};
    int resultIdx = 0;
    int numIdx = 0;
    
    for (int i = 0; encoded[i]; i++) {
        if (isdigit(encoded[i])) {
            currentNum[numIdx++] = encoded[i];
        } else if (isalpha(encoded[i])) {
            if (numIdx > 0) {
                currentNum[numIdx] = '\0';
                int count = atoi(currentNum);
                for (int j = 0; j < count; j++) {
                    result[resultIdx++] = encoded[i];
                }
                numIdx = 0;
                memset(currentNum, 0, sizeof(currentNum));
            } else {
                result[0] = '\0';
                return result;  // Invalid format
            }
        } else {
            result[0] = '\0';
            return result;  // Invalid character
        }
    }
    
    // Check if we ended with a number (invalid)
    if (numIdx > 0) {
        result[0] = '\0';
        return result;
    }
    
    result[resultIdx] = '\0';
    return result;
}

int main() {
    char encoded[1000];
    fgets(encoded, sizeof(encoded), stdin);
    encoded[strcspn(encoded, "\n")] = 0;  // Remove newline
    
    char* result = solution(encoded);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + m)

n is input length, m is output length (sum of all counts)

✓ Linear Growth

Space Complexity

O(m)

Output string stores m characters where m is sum of all counts

✓ Linear Space

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Run-Length Decoder - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Character-by-Character Parsing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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