Minimizing time complexity when looping over two arrays - c

Question

For a intro computer science class we have to write a function that finds the intersection of two arrays, which each unique element only being shown once and without allocating more space than we need.
For example:
array A = {1, 2, 3, 3, 3, 5}
array B = {2, 2, 2, 3, 5, 6}
intersection of A and B = {2, 3, 5}
How can I accomplish this without looping over the both arrays twice? As it stands I have:
//find how large of an array I'll need
for array A
for array A
if A[i] is already somewhere earlier in array A
stop
else
loop through array B
if A[i] is in array B, increment a counter
declare a new array of size counter
//add unique elements to the array
for array A
for array A
if A[i] is already somewhere earlier in array A
stop
else
loop through array B
if A[i] is in array B, add it to the new array
It seems like this would be really inefficient I have two nearly identical nested for loops. If I was using python I could just append unique elements to list, but is there a way I could do something similar in C? I could just declare an array of the maximum size I could need, but I'm trying to minimize the space complexity.

Answer 1

If your are aware of sets you can use that. The time-complexity will be O(n)
You can know more about set and how to implement one in C here.
Then you can do something like this (written in Java):
public int[] intersection(int[] nums1, int[] nums2) {
Set<Integer> set1 = getSet(nums1);
Set<Integer> set2 = getSet(nums2);
Set<Integer> ans = new HashSet<>();
for(Integer i: set1) {
if(set2.contains(i)) {
ans.add(i);
}
}
int[] ret = new int[ans.size()];
int count = 0;
for(Integer i: ans) {
ret[count] = i;
count++;
}
return ret;
}
public Set<Integer> getSet(int[] arr) {
Set<Integer> set = new HashSet<>();
for(int a: arr) { set.add(a); }
return set;
}