CareerCup

simply use a hash table and solve it O(n+m).
int[] union
for i=1 to n
if(H.find(A[i])!=true)
H.put(A[i]),union.append(A[i]);
end
end
for i=1 to m
if(H.find(B[i])!=true)
H.put(B[i]),union.append(B[i]);

simply use a hash table and solve it O(n+m).
int[] union
for i=1 to n
if(H.find(A[i])!=true)
H.put(A[i]),union.append(A[i]);
end
end
for i=1 to m
if(H.find(B[i])!=true)
H.put(B[i]),union.append(B[i]);
return union

simply use a hash table and solve it O(n+m).
int[] union
for i=1 to n
if(H.find(A[i])!=true)
H.put(A[i]),union.append(A[i]);
end
end
for i=1 to m
if(H.find(B[i])!=true)
H.put(B[i]),union.append(B[i]);
end
end
return union

Make a BST of set A's elements and then try to find B's elements in it. If any of B's element is not found then insert it. Thus found BST is (A U B).

Sort the sets if you are not able to have an hash table.
Them merge them if you get equal numbers eliminate one of them in the new set. That is all.

BST is a good way

Use rooted tree and solve it in O(n+m).

Simple use another HashSet and keep adding the element of both the array, At the end you will have all the elements of both the array without repetition.
Time Complexity O(max(n,m)), Space O(n+m)