The difference between 2 arrays in c++ A\B - c++

Question

So I have 2 arrays. Let's say the first one it's called a and the second one b. The first one uses "i" for it's elements and the second one uses "j".
For example we have a[ 1 2 3 4] and b[3 4 5] it should show c[1 2]. In the array c I want to show the elements that are in a and aren't in b.
This is what I've tried, but without succes:
#include <iostream>
using namespace std;
int main(int argc, char const *argv[]) {
int a[50], b[50], c[50], i, j, k, n, m;
cout << "n= "; cin >> n;
//Read arrays
for (i = 0; i < n; i++) {
cout << "a[" << i << "]: "; cin >> a[i];
}
cout << "\nm= "; cin >> m;
for (j = 0; j < m; j++) {
cout << "b[" << j << "]: "; cin >> b[j];
}
//Show the arrays
cout << endl;
cout << "\na[ ";
for (i = 0; i < n; i++) {
cout << a[i] << " ";
}
cout << "]";
cout << endl;
cout << "\nb[ ";
for (j = 0; j < m; j++) {
cout << b[j] << " ";
}
cout << "]";
//Calculate the difference
k = 0; i = 0;
for (j = 0; j < m; j++) {
if (a[i] != b[j])
c[k] = a[i];
k++;
while (j == m && i < n)
i++;
}
//Show the difference array
cout << endl;
cout << "\nc[ ";
for (i = 0; i < k; i++) {
cout << c[i] << " ";
}
cout << "]";
return 0;
}

Answer 1

If the items are sorted, use std::set_difference:
#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>
int main()
{
int a[] = { 1, 2, 3, 4 };
int b[] = { 3, 4, 5 };
std::vector<int> cv;
std::set_difference(std::begin(a), std::end(a),
std::begin(b), std::end(b),
std::back_inserter(cv));
for (auto& s : cv)
std::cout << s << "\n";
}
Output:
1
2
The advantage of using the STL algorithms is that the purpose of the code is known immediately just by looking at the name of the function, and that they work every time (if you give them the correct parameters). Note the lack of comments -- any competent C++ programmer understands right away what's being done.
On the other hand, if you didn't mention what your original code was trying to do (including removing the comments), it would take much more effort to figure out what it's supposed to be doing, and as you've seen, it contains bugs.

Answer 2

Your logic is wrong.
Explanation
So the thing that we will do
For each element in a we will have to check if it is there in array b or not.
If we see any element of a[i] in b[1..m] then we can't add it to c.
So in code we just mark it by f=1
When I get out of that second for loop I want to check if that a[i] is eqaul to any of the element in b[1..m] in which case f will be 1. But if it is 0 then add it to array c[].
Correct one
int k=0;
for(int i=0;i<n;i++)
{
int f=0;
for(int j=0;j<m;j++)
if(a[i]==b[j])
f=1;
if(!f)
c[k++]=a[i];
}
Where OP went wrong?
Being not equal to one element of b[] doesn't guarantee that the element is not appearing b[0..m-1] . This is where op went wrong.
In the for loop
for(j=0;j<m;j++) you are checking if particular a[i] is equal to b[j] or not. If that is the case then it is added to c[] . It is wrong. Also i is not incremented in the loop unless j==m and as in the for loop the condition is j<m so i is never incremented. And k is incremented every time so not every element in c is valid they may contain garbage value even after processing.
k = 0; i = 0;
for (j = 0; j < m; j++) {
if (a[i] != b[j]) // this doesn't mean that it is not appearing in `b`
c[k] = a[i];
k++; // k is incremented in every iteration which is wrong. It should be only when we are sure that `a[i]` is not in `b[0..m-1] `
while (j == m && i < n)
i++; // OP is not using it anywhere...this is redundant.
}
what op did?
Compared first element of a[0] with every element of b[0..m-1] and array c[] contains m elements irrespective of what a[] and b[] is, out of which
c[i]={ a[0] if b[j]==a[0]
{ garbage value if b[j] not equal to a[0]
Dry Run of OP's code
k = 0; i = 0;
for (j = 0; j < m; j++) {
if (a[i] != b[j])
c[k] = a[i];
k++;
while (j == m && i < n)
i++;
}
Input
Case: 1 2 3 4 :a[]
2 3 4 1 :b[]
Step 1: i=0 a[0]!=b[0] is true so c[0]=a[0]. the `while loop` not entered.
j++
Step-2: i is still 0. a[0]!=b[1] so it is added c[1]=a[0]. While loop not entered.
j++
Step-3: i is still 0. a[0]!=b[2]. So c[2]=a[0]. While loop skipped.
j++
Step-4: i is still 0. a[0]==b[3] is true so no assignment done. But k is incremented. so c[3]=garbage. j=3 so while loop skipped
j++
Out of for loop.
Output: [here x is garbage value]
a[]: 1 2 3 4
b[]: 2 3 4 1
c[]: 1 2 3 x
Example test case
1 2 3 4 :=a
2 3 4 1 :=b
Corrected Code
#include <iostream>
using namespace std;
int main(int argc, char const *argv[]) {
int a[50], b[50], c[50], i, j, k, n, m;
cout << "n= "; cin >> n;
//Read arrays
for (i = 0; i < n; i++) {
cout << "a[" << i << "]: "; cin >> a[i];
}
cout << "\nm= "; cin >> m;
for (j = 0; j < m; j++) {
cout << "b[" << j << "]: "; cin >> b[j];
}
//Show the arrays
cout << endl;
cout << "\na[ ";
for (i = 0; i < n; i++) {
cout << a[i] << " ";
}
cout << "]";
cout << endl;
cout << "\nb[ ";
for (j = 0; j < m; j++) {
cout << b[j] << " ";
}
cout << "]";
//Calculate the difference
k = 0; i = 0;
int k=0;
for(int i=0;i<n;i++)
{
int f=0;
for(int j=0;j<m;j++)
{
if(a[i]==b[j])
f=1;
if(!f)
c[k++]=a[i];
}
}
//Show the difference array
cout << endl;
cout << "\nc[ ";
for (i = 0; i < k; i++) {
cout << c[i] << " ";
}
cout << "]";
return 0;
}

Answer 3

Your code seems to check if the elements in a are equal to all elements of b. If you just want to check the elements in a if they are equal to at least one element of b, you can do
for (int i=0; i<n; i++) {
bool found = false;
for (int j=0; j<m; j++) {
if (a[i] == b[j]) {
found = true;
break;
}
}
if (!found) {
std::cout << "a["<<i<<"] is not in b"<<std::endl;
}
}
Or add the element to c, but I would recommend to use std::vector<int> c for that.

Answer 4

In the array c I want to show the elements that are in a and aren't in b
It seems like you are looking for std::set_difference
int a[4] = {1, 2, 3, 4}, b[3] = {3, 4, 5};
int c[2] = {}; // declare c with enough space to hold all the elements in result
std::set_difference(a, a + 4, b, b + 3, c); // now c contains the element that are in a but not in b

Answer 5

You can do this very easily using 'set'.
#include<iostream>
#include<set>
int main(){
std::set<int> a = {1,2,3,4} , b = {3,4,5};
for(int const inB : b)
a.erase(inB);
for(int const inA : a)
std::cout << inA << " ";
std::cout << std::endl;
return 0;
}