Apply Operations to an Array | LeetCode 2460 | C

2023-08-29
#LeetCode #C #Array
LeetCode

You are given a 0-indexed array nums of size n consisting of non-negative integers.

You need to apply n - 1 operations to this array where, in the ith operation (0-indexed), you will apply the following on the ith element of nums:

If nums[i] == nums[i + 1], then multiply nums[i] by 2 and set nums[i + 1] to 0. Otherwise, you skip this operation.
After performing all the operations, shift all the 0’s to the end of the array.

For example, the array [1,0,2,0,0,1] after shifting all its 0’s to the end, is [1,2,1,0,0,0].
Return the resulting array.

Note that the operations are applied sequentially, not all at once.

Example 1:

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Input: nums = [1,2,2,1,1,0]
Output: [1,4,2,0,0,0]
Explanation: We do the following operations:
- i = 0: nums[0] and nums[1] are not equal, so we skip this operation.
- i = 1: nums[1] and nums[2] are equal, we multiply nums[1] by 2 and change nums[2] to 0. The array becomes [1,4,0,1,1,0].
- i = 2: nums[2] and nums[3] are not equal, so we skip this operation.
- i = 3: nums[3] and nums[4] are equal, we multiply nums[3] by 2 and change nums[4] to 0. The array becomes [1,4,0,2,0,0].
- i = 4: nums[4] and nums[5] are equal, we multiply nums[4] by 2 and change nums[5] to 0. The array becomes [1,4,0,2,0,0].
After that, we shift the 0's to the end, which gives the array [1,4,2,0,0,0].

Example 2:

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Input: nums = [0,1]
Output: [1,0]
Explanation: No operation can be applied, we just shift the 0 to the end.

Constraints:

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2 <= nums.length <= 2000
0 <= nums[i] <= 1000

Approach

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Just follow the instructed operation and go through the integer array.

Algorithm

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Go through the arry.
Find out if nums[i] == nums[i + 1] and do the respective operation.
Mind the last element in the original array.

Implementation

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/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* applyOperations(int* nums, int numsSize, int* returnSize){
    int * result = calloc(numsSize,sizeof(int));
    int index = 0;
    for(int i;i<numsSize-1;i++)
    {
        if(nums[i]==nums[i+1])
        {
            nums[i] = nums[i]*2;
            nums[i+1] = 0;
            if(nums[i]!=0)
            {
                result[index] = nums[i];
                index++;
            }
        }
        else{
            if(nums[i]!=0)
            {
                result[index] = nums[i];
                index++;
            }
            continue;
        }
    }
    if (nums[numsSize-1]!=0)
    {
        result[index] = nums[numsSize-1];
    }
    *returnSize = numsSize;

    return result;
}