# Problem in “between two set” hackerrank problem

#### Tags: algorithm, java

Problem:You will be given two arrays of integers and asked to determine all integers that satisfy the following two conditions:

The elements of the first array are all factors of the integer being considered The integer being considered is a factor of all elements of the second array These numbers are referred to as being between the two arrays. You must determine how many such numbers exist.

for example :Sample Input

```2 3
2 4
16 32 96
```

Sample Output

```3
```

My code:

```public static int getTotalX(int n, int m, List<Integer> a, List<Integer> b) {
int total=0,x=0,y=0;
for(int i=a.get(n-1);i<=b.get(0);i++)
{
for(int j=0;j<n;j++)
{
//to check if the elements in list 'a' can divide the integer.
if(i%a.get(j)==0)
{
y++;
}
}
//if every element in list a can divide the integer go forward
if(y==n)
{
for(int k=0;k<m;k++)
{
//to check if the elements of list 'b' is divisible by integer
if(b.get(k)%i==0)
{
x++;
}
}
y=0;
//if every element of 'b' is divisible by integer, count how many                        such integers are there
if(x==m)
{
total++;
x=0;
}
}
}

}
```

My code is not giving proper solution and I cant understand why.

```private static int getTotalX(int n, int m, List<Integer> a, List<Integer> b) {
int total = 0, x = 0, y = 0;
for (int i = a.get(n - 1); i <= b.get(0); i++) {
for (int j = 0; j < n; j++) {
if (i % a.get(j) == 0) {
y++;
}
}
if (y == n) {
for (int k = 0; k < m; k++) {
if (b.get(k) % i == 0) {
x++;
}
}
if (x == m) {
total++;
}
}
// changes here
y = 0;
x = 0;
}
}
```

Great progress. You were very close. Algorithm is on point and efficient.

Only one mistake: you were resetting the variables `x` and `y` inside the `if` conditions.

What if the condition is not true? Then the variables are never reset and all the future computations are done on those wrong values in `x` and `y`.

Like Java8? Here is a one-liner:

```return (int) IntStream.rangeClosed(a.get(n - 1), b.get(0))
.filter(i -> a.stream().filter(value -> i % value == 0).count() == a.size())
.filter(i -> b.stream().filter(value -> value % i == 0).count() == b.size())
.count();
```

Source: stackoverflow