Common elements within 2 arrays - arrays

Question

I'm trying to write a function that copies all of the values in source1 which are also found in source2 into a destination and then returns the number of elements copied into the destination.
int common_elements(int length, int source1[length], int source2[length], int destination[length])
{
int counter = 0;
int i = 0;
while (i < length) {
int j = 0;
while (j < length) {
if ( source1[i] == source2[j]) {
destination[counter] = source1[i];
counter++;
}
j++;
}
i++;
}
return counter;
}
The problem is e.g. given (common_elements(5, {1,2,3,4,5}, {1,2,3,2,1}, [])), the correct input should be
1,2,3
return value: 3
However, the program is accounting for the duplicates and produces :
1,1,2,2,3
return value: 5
which is incorrect.
How can I remedy this?

Answer 1

In this while loop
int j = 0;
while (j < length) {
if ( source1[i] == source2[j]) {
destination[counter] = source1[i];
counter++;
}
j++;
}
you are counting all elements in the array source2 that are equal to the element source1[i].
I can suggest the following solution provided that the source arrays may not be changed within the function.
#include <stdio.h>
size_t common_elements( int destination[],
const int source1[],
const int source2[],
size_t n )
{
size_t counter = 0;
for ( size_t i = 0; i < n; i++ )
{
size_t number = 1;
for ( size_t j = 0; j < i; j++ )
{
if ( source1[i] == source1[j] ) ++number;
}
for ( size_t j = 0; number && j < n; j++ )
{
if ( source1[i] == source2[j] ) --number;
}
if ( number == 0 ) destination[counter++] = source1[i];
}
return counter;
}
int main(void)
{
enum { N = 5 };
int source1[N] = { 1, 2, 3, 4, 5 };
int source2[N] = { 1, 2, 3, 2, 1 };
int destination[N];
size_t n = common_elements( destination, source1, source2, N );
for ( size_t i = 0; i < n; i++ )
{
printf( "%d ", destination[i] );
}
putchar( '\n' );
return 0;
}
The program output is
1 2 3