Decompress Run-Length Encoded List

Decompress Run-Length Encoded List - Problem

Array Easy

We are given a list nums of integers representing a list compressed with run-length encoding.

Consider each adjacent pair of elements [freq, val] = [nums[2*i], nums[2*i+1]] (with i >= 0). For each such pair, there are freq elements with value val concatenated in a sublist. Concatenate all the sublists from left to right to generate the decompressed list.

Return the decompressed list.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3,4]

› Output: [2,4,4,4]

💡 Note: First pair [1,2]: frequency 1, value 2 → add 2 once. Second pair [3,4]: frequency 3, value 4 → add 4 three times. Result: [2,4,4,4]

Example 2 — Multiple Pairs

$ Input: nums = [1,1,2,3]

› Output: [1,3,3]

💡 Note: First pair [1,1]: frequency 1, value 1 → add 1 once. Second pair [2,3]: frequency 2, value 3 → add 3 twice. Result: [1,3,3]

Example 3 — Zero Frequency

$ Input: nums = [2,1,3,2]

› Output: [1,1,2,2,2]

💡 Note: First pair [2,1]: frequency 2, value 1 → add 1 twice. Second pair [3,2]: frequency 3, value 2 → add 2 three times. Result: [1,1,2,2,2]

Constraints

2 ≤ nums.length ≤ 100
nums.length % 2 == 0
1 ≤ nums[2 * i] ≤ 100
1 ≤ nums[2 * i + 1] ≤ 100

Visualization

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Asked in

G Google 15 a Amazon 12 f Facebook 8 M Microsoft 6

The key insight is that run-length encoding stores data as [frequency, value] pairs, and we need to expand each pair into multiple elements. The optimal approach processes each pair once, expanding values directly into the result array. Time: O(n), Space: O(n) where n is the total output size.

Common Approaches

✓ Direct Expansion

⏱️ Time: O(n) Space: O(n)

Iterate through the input array in steps of 2, treating each pair as [frequency, value]. For each pair, append the value to the result array exactly frequency times.

Single Pass with Pre-allocation

⏱️ Time: O(n + m) Space: O(m)

First pass calculates the total size needed for the result array, then pre-allocate the array and fill it in a second pass. This avoids dynamic resizing of the result array.

Direct Expansion — Algorithm Steps

Step 1: Initialize empty result array
Step 2: Process pairs at indices (0,1), (2,3), (4,5)...
Step 3: For each pair [freq, val], append val to result freq times

Visualization

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Step-by-Step Walkthrough

Input Parsing

Read pairs from input array

Expansion

For each pair, repeat value frequency times

Concatenation

Combine all expanded elements

Code -

solution.c — C

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

int* solution(int* nums, int numsSize, int* returnSize) {
    int capacity = 1000;
    int* result = (int*)malloc(capacity * sizeof(int));
    int size = 0;
    
    for (int i = 0; i < numsSize; i += 2) {
        int freq = nums[i];
        int val = nums[i + 1];
        for (int j = 0; j < freq; j++) {
            if (size >= capacity) {
                capacity *= 2;
                result = (int*)realloc(result, capacity * sizeof(int));
            }
            result[size++] = val;
        }
    }
    
    *returnSize = size;
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int returnSize;
    int* result = solution(nums, numsSize, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We process each element in input once and output each element once, where n is total output size

✓ Linear Growth

Space Complexity

O(n)

Result array stores all decompressed elements

⚡ Linearithmic Space

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Medium Frequency

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Input & Output

Constraints

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Related Problems

Common Approaches

Direct Expansion — Algorithm Steps

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Code -

Time & Space Complexity

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