LeetCode //C - 1313. Decompress Run-Length Encoded List

1313. Decompress Run-Length Encoded List

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, nums2*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.

Example 1:

Input: nums = 1,2,3,4

Output: 2,4,4,4

Explanation: The first pair 1,2 means we have freq = 1 and val = 2 so we generate the array 2.

The second pair 3,4 means we have freq = 3 and val = 4 so we generate 4,4,4.

At the end the concatenation 2 + 4,4,4 is 2,4,4,4.

Example 2:

Input: nums = 1,1,2,3

Output: 1,3,3

Constraints:

• 2 <= nums.length <= 100
• nums.length % 2 == 0
• 1 <= numsi <= 100

From: LeetCode

Link: 1313. Decompress Run-Length Encoded List

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Solution:

Ideas:

First calculate the total output size, allocate the array, then insert each val exactly freq times.

Code:

#include <stdlib.h>

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* decompressRLElist(int* nums, int numsSize, int* returnSize) {
    int totalSize = 0;

    // Calculate the decompressed array size.
    for (int i = 0; i < numsSize; i += 2) {
        totalSize += nums[i];
    }

    int* result = (int*)malloc(totalSize * sizeof(int));
    int index = 0;

    // Add each value frequency times.
    for (int i = 0; i < numsSize; i += 2) {
        int frequency = nums[i];
        int value = nums[i + 1];

        for (int j = 0; j < frequency; j++) {
            result[index++] = value;
        }
    }

    *returnSize = totalSize;
    return result;
}