Implement operators for - and -= for the bag class. For two bags
x and y, the bag x-y contains all the items of x, with any items
from y removed. For example,
suppose that x has seven copies of the number 3, and y has two
copies of the number 3. Then x-y will have five copies of the
number 3 (i.e., 7 - 2 copies of the number 3). In the case where y
has more copies of an item than x does, the bag x-y will have no
copies of that item. For example, suppose that x has nine copies of
the number 8, and y has 10
copies of the number 8. Then x-y will have no 8s. The statement x
-= y should have the same effect as the assignment x = x-y;
I tried one program same from Chegg solutions but it didn't work can you please help me? debug so it will work the complete program in visual basic use c++ language.
PLEASE GIVE THUMBS UP, THANKS
Sample output:

code:
#include<iostream>
#include<vector>
using namespace std;
class Bag
{
private:
vector<int> List;
public:
Bag()
{
List=vector<int>();
}
void insert(int data)
{
List.push_back(data);
}
int getSize(){
return
List.size();
}
int getElementAt(int i)
{
return
List.at(i);
}
void setElementAt(int i,int
data)
{
List.at(i)=data;
}
void deleteElementAt(int
index)
{
List.erase(List.begin()+index);
}
Bag operator -(Bag &B)
{
Bag
temp1=Bag();
Bag
temp2=Bag();
for(int i=0;
i<this->getSize();i++)
{
temp1.insert(this->getElementAt(i));
}
for(int i=0;
i<B.getSize();i++)
{
temp2.insert(B.getElementAt(i));
}
for(int
i=temp2.getSize()-1;i>=0; i--)
{
for(int j=temp1.getSize()-1; j>=0; j--)
{
if(temp1.getElementAt(j)==temp2.getElementAt(i))
{
temp1.deleteElementAt(j);
temp2.getElementAt(i);
break;
}
}
}
return
temp1;
}
void print()
{
for(int i=0;
i<List.size();i++)
{
cout<<List.at(i)<<" ";
}
cout<<endl;
}
};
int main()
{
Bag A=Bag();
Bag B=Bag();
for(int i=1; i<=10; i++)
{
A.insert(8);
}
cout<<"Bag A : ";A.print();
for(int i=1; i<=7; i++)
{
B.insert(8);
}
cout<<"Bag B : ";B.print();
A=A-B;
cout<<"Print A=A-B : ";A.print();
}
Implement operators for - and -= for the bag class. For two bags x and y,...
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