Intersection of 2 dynamically allocated arrays c++ - c++

Question

I am trying to create a function that will find the intersection of two dynamically allocated arrays comparing array 1 to array 2. For any values in array 1 that are not in array 2, those values should be deleted in array 1 so that array 1 now only holds the common values of both arrays (no repeats). I cannot use vectors, hashes, or any other thing outside of my current functions in my class:
here is my code so far:
bool IntSet::contains(int val) const
{
for (int i = 0; i < numValues; i++)
{
if (set[i] == val)
return true;
}
return false;
}
this function compares an integer parameter to values currently stored in the array...if a value is in the array it returns true and if else false;
this next function takes in a value and removes that value from the array:
void IntSet::remove(int val)
{
for (int i = 0; i < numValues; i++)
{
if (set[i] == val)
for (int j = 0; j < numValues; j++)
set[j] = set[j + 1];
}
numValues--;
}
here's where I've been having problems, this next function is supposed to iterate through one array and compare those values with the values in the other array...if one value from one array is in the other, it should just skip it, but if a value is not in the array calling the function, it should delete that value from the calling array:
void IntSet::removeDifferent(const IntSet &set2)
{
for (int i = 0; i < set2.size(); i++)
{
if (!set2.contains(set[i]))
{
remove(set[i]);
}
}
}
ive tried about 50 different variations on the removeDifferent() function and I just can't seem to figure this one out. Could someone point me in the right direction?

Answer 1

You're iterating i through the indexes of set2, but then you're testing set[i]. Try this:
void IntSet::removeDifferent(const IntSet &set2)
{
for (int i = 0; i < numValues; ) {
if (!set2.contains(set[i])) {
remove(set[i]);
} else {
i++;
}
}
Note that I also removed i++ from the for loop header. This is because when you remove an element, all the following elements are shifted down, so the next element takes its place in the array. If you incremented i, it would skip that element.
You also need to fix remove. It should start its inner loop from i, so it only shifts down the elements after the one being removed, and it should stop at numValues-1, so it doesn't try to access outside the array when it copies set[j+1]. And as an optimization, it can break out of the outer loop once it has found a match (I assume IntSet doesn't allow duplicates, since you only decrement numValues by 1).
void IntSet::remove(int val)
{
for (int i = 0; i < numValues; i++)
{
if (set[i] == val) {
for (int j = i; j < numValues - 1; j++) {
set[j] = set[j + 1];
}
break;
}
}
numValues--;
}

Answer 2

Your problem is in your remove() function:
void IntSet::remove(int val)
{
for (int i = 0; i < numValues; i++)
{
if (set[i] == val)
for (int j = 0; j < numValues; j++)
set[j] = set[j + 1];
}
numValues--;
}
You can figure out yourself why this is wrong by using a paper and pencil here. Start with a typical example: let's say you found the value you're looking for in the third element of a five-element array:
if (set[i] == val)
In this example, i would be set to 2, and numValues would be set to five. It doesn't matter what val is. Whatever it is, you found it when i is 2, and numValues is five: you found it in the third element of a five element array. Keep that in mind.
Now, you know that you are now supposed to remove the third element in this five element array. But what do you think will happen next:
for (int j = 0; j < numValues; j++)
set[j] = set[j + 1];
Well, using the aforementioned paper and pencil, if you work it out, the following will happen:
set[1] will be copied to set[0]
set[2] will be copied to set[1]
set[3] will be copied to set[2]
set[4] will be copied to set[3]
set[5] will be copied to set[4]
There are two problems here:
A) There is no set[5]. Recall that this is a five-element array, si you only have set[0] through set[4]
B) You're not supposed to copy everything in array down to one element. You have to copy only the elements after the element you want to remove.
Fix these two problems, and you will probably find that everything will work correctly.