How to identify an embedded palindrome within an array of digits - arrays

Question

Say we have
int arr1[]={1,2,3,4,5};
int temp1[], temp2[];
and that we need to copy the first 3 members from arr1 to temp1, and the second 3 from arr1 to temp2, so that
temp1={1,2,3};
temp2={2,3,4};
and so on and so forth, how would you go on about doing it? I am trying to write a program to check if a user inputted number contains a palindrome in a length k (where k is also chosen by the user) and my idea was to store the number in an array, take the first 3 members, put them in an array, reverse them and put that in another array and then compare, but I am stuck on how to solve the problem I mentioned above, I tried something like:
void copyandReverse(int arr[], int copy[], int reverse[], int start, int length)
{
for(int i=0; i<length; i++)
{
copy[start+i]=arr[start+i];
reverse[start+i]=arr[start+length-i-1];
}
}
but it seems to only copy the first 3 elements, and reverse them.
PS: I don't think we're allowed to use string or dynamic memory allocation, it's given that the inputted number contains less then 10 digits, so I made the temporary arrays with a constant size 10.

Answer 1

There is no any need to create auxiliary arrays to check whether a given array is a palindrome.
This can be done much simpler.
Write a function like this
int is_palindrome( const unsigned int a[], size_t n )
{
size_t i = 0;
while (i < n / 2 && a[i] == a[n - i - 1]) ++i;
return i == n / 2;
}
then in main if you have an array like
unsigned int a[10];
and you need to check whether a sub-array with three elements is a palindrome then you can do this in a for loop
size_t k = 3;
for ( size_t i = 0; i <= 10 - k; i++ )
{
if ( is_palindrome( a + i, k ) )
{
printf( "The sub-array at position %zu:", i );
for ( size_t j = i; j < k + i; j++ ) printf( " %u", a[j] );
puts( " is a palindrome" );
}
}

Answer 2

as #stark mentioned, just iterate over the array with two indices.
Please see an example implementation below. You just need to evaluate the actual size from the userinput before you validate it.
The code below assumes 0 as the delimited. After gathering the actual user input size, we start a loop with index i and j. Looping until they have swapped positions.
If a mismatch exists, we return FALSE, otherwise TRUE.
#include <stdio.h>
#define TRUE 1
#define FALSE 0
size_t getUserInputLength(int *arr, size_t len)
{
for (size_t i = 0; i < len; i++)
{
if (arr[i] == 0)
{
return i;
}
}
return len;
}
int checkPalindrome(int *arr, size_t len)
{
size_t i, j;
for (i = 0, j = len - 1; i <= j; i++, j--)
{
printf("Current indices: [i: %ld, j: %ld]\n", i, j);
printf("Current values: [i: %d, j: %d]\n", arr[i], arr[j]);
if (arr[i] != arr[j])
{
return FALSE;
}
}
return TRUE;
}
int main()
{
// even number of user input
int evenInput[] = {1, 2, 3, 4, 3, 2, 1, 0, 0, 0};
size_t evenLen = getUserInputLength(evenInput, 10);
printf("Even number returned: %d\n", checkPalindrome(evenInput, evenLen));
// odd number of user input
int oddInput[] = {1, 2, 3, 2, 1, 0, 0, 0, 0, 0};
size_t oddLen = getUserInputLength(oddInput, 10);
printf("Odd number returned: %d\n", checkPalindrome(oddInput, oddLen));
return 0;
}