Deleting Elements from an Array based on their Position every 3 Elements - c

Question

I created an array called elements_n which has the elements 0 to N-1 where N is 2. The below numbers are the elements of the array called elements_n:
0 1
I have another array called Arr which has the following elements:
0 1 3 1 2 4
If any of the first 3 elements of the array Arr are equal to the first element of elements_n which is 0, I would like to delete that element from the array called Arr. I then repeat the same process for the next 3 elements of the array Arr. So to explain myself better, I will use the following example:
Compare the first 3 elements of array Arr which are 0, 1, 3 to the first element of elements_n which is 0. Since Arr[0] == elements_n[0]. I delete Arr[0] from the array Arr.
Compare the next 3 elements of array Arr which are 1, 2, 4 to the second element of elements_n which is 1. Since Arr[3] == elements_n[1]. I delete Arr[3] from the array Arr. So the elements that should be left in the array Arr are:
1 3 2 4
When I implemented it myself in C programming with the code found below the end result is coming:
1 3 3 2 2 4
Rather than:
1 3 2 4
This is the code I implemented:
#include <stdio.h>
#include <stdlib.h>
#define N 2
int main() {
unsigned *elements_n = malloc(N * sizeof(unsigned));
for (int i = 0; i < N; i++) {
elements_n[i] = i; //Created an array which has the elements 0 to N-1
}
printf("\n");
unsigned Arr[6] = { 0, 1, 3, 1, 2, 4 };
unsigned position_indices[2] = { 3, 3 }; //Moving every 3 elements in the Arr array.
int count = 0;
int index = 0;
unsigned *ptr_Arr = &Arr[0];
do {
for (int i = 0; i < position_indices[count]; i++) {
if (ptr_Arr[i] == elements_n[count]) {
index = i + 1; //Index of the Arr element that has the same value as the element in the array elements_n
for (int j = index - 1; j < position_indices[count] - 1; j++) {
ptr_Arr[j] = ptr_Arr[j + 1];
}
}
}
printf("\n");
ptr_Arr += position_indices[count] - 1;
count++;
} while (count < 2);
for (int i = 0; i < 6; i++) {
printf("%d\t", Arr[i]);
}
printf("\n");
free(elements_n);
return 0;
}

Answer 1

You might try something like this (not tested).
#include <stdio.h>
#include <stdlib.h>
#define N 2
int main()
{
unsigned *elements_n = malloc(N * sizeof(unsigned));
for (int i = 0; i < N; i++)
{
elements_n[i] = i; //Created an array which has the elements 0 to N-1
}
unsigned Arr[6] = { 0, 1, 3, 1, 2, 4 };
int dest_index = 0;
int src_index = 0;
int count = sizeof(Arr)/sizeof(Arr[0]);
for ( ; src_index < count; src_index++)
{
int group = src_index / 3;
if (Arr[src_index] != elements_n[group])
{
Arr[dest_index++] = Arr[src_index];
}
}
for (int i = 0; i < dest_index; i++)
{
printf("%d\t", Arr[i]);
}
printf("\n");
free(elements_n);
return 0;
}

Answer 2

You need to keep track of how many elements you removed from the array.
My solution:
#include <stdio.h>
#include <stddef.h>
#include <assert.h>
#include <string.h>
size_t fancy_delete_3(const int elems[], size_t elemssize, int arr[], size_t arrsize)
{
assert(elems != NULL);
assert(arr != NULL);
assert(arrsize%3 == 0);
assert(elemssize*3 == arrsize);
// we need to count the removed elements, to know how much we need to shift left
size_t removed = 0;
// for each element in elems
for (size_t i = 0; i < elemssize; ++i) {
// check the three correponding elements in arr
for (size_t j = i*3; j < (i+1)*3; ++j) {
assert(j >= removed);
const size_t pos = j - removed;
// if elems[i] matches any of the corresponding element in arr
if (elems[i] == arr[pos]) {
// remove element at position pos
assert(arrsize >= pos + 1);
// I don't think this can ever overflow
memmove(&arr[pos], &arr[pos + 1], (arrsize - pos - 1) * sizeof(int));
++removed;
// array is one element shorter, so we can just decrease the array size
assert(arrsize > 0);
--arrsize;
}
}
}
// we return the new size of the array
return arrsize;
}
#define __arraycount(x) sizeof(x)/sizeof(x[0])
int main()
{
int elements_n[] = {0,1};
int arr[] = {0,1,3, 1,2,4};
size_t newsize = fancy_delete_3(elements_n, __arraycount(elements_n), arr, __arraycount(arr));
printf("arr size=%zu {", newsize);
for (size_t i = 0; i < newsize; ++i)
printf("%d,", arr[i]);
printf("}\n");
return 0;
}

Answer 3

You have several related problems around how you perform deletions. In the first place, it's not clear that you understand that you cannot actually delete anything from a C array. The closest you can come is to overwrite it with something else. Often, pseudo-deletion from an array is implemented by moving each of the elements following the deleted one one position forward, and reducing the logical length of the array.* You seem to have chosen this alternative, but (problem 1) you miss maintaining or updating a logical array length.
Your problem is made a bit more complicated by the fact that you logically subdivide your array into segments, and you seem not to appreciate that your segments are variable-length in that, as described, they shrink when you delete an element. This follows from the fact that deleting an element from one group does not change the assignments of elements to other groups. You do have a mechanism in position_groups that apparently serves to track the sizes of the groups, and in that sense its name seems ill-fitting. In the same way that you need to track and update the logical length of the overall array, you'll need to track and update the lengths of the groups.
Finally, you appear to have an off-by-one error here:
for (int j = index - 1; j < position_indices[count]-1; j++)
that would be clearer if position_indices were better named (see above), but recognizing that what it actually contains is the size of each group, and that index and j represent indices within the group, it follows that the boundary condition for the iteration should instead be just j < position_indices[count]. That's moot, however, because you're going to need a somewhat different approach here anyway.
Suggestion, then:
When you delete an element from a group, move up the entire tail of the array, not just the tail of the group.
In service to that, update both group size and logical array size when you perform a deletion, remembering that that affects also where each subsequent group starts.
When you examine or output the result, remember to disregard array elements past the logical end of the array.
* "Logical array size" means the number of (leading) elements that contain meaningful data. In your case, the logical array size is initially the same as the physical array size, but each time you delete an element (and therefore move up the tail) you reduce the logical size by one.

Answer 4

You never really "deleted" any element, you just shifted-down the
second and third elements from each 3-group. And then, when
printing, you iterated over the whole array.
Arrays are just continuous blocks of memory, so you need a
different strategy. You could traverse the original array with
two indexes, one as the general index counter and other as
the index for the shifted slot. If the current element is
different from the corresponding elements_n item, copy it to
the secondary index and increase it.
In case nothing is equal to the elements_n item, you would just
reassign the array elements to themselves. However, as soon as
one is equal you will be shifting them with the advantage of
keeping track of the new size.
Also, calculating the corresponding elements_n item is a simple
matter of dividing the current index by 3, so you don't even need
an extra variable to that.
#include <stdio.h>
#define N 2
int main()
{
unsigned elements_n[N] = { 0, 1 };
unsigned Arr[N*3] = { 0, 1, 3, 1, 2, 4 };
int i, j;
for (i = 0, j = 0; i < N*3; i++)
if (Arr[i] != elements_n[i/3])
Arr[j++] = Arr[i];
for (i = 0; i < j; i++)
printf(" %d", Arr[i]);
printf("\n");
return 0;
}