Two pointers means that in order to solve the problem, there is an invariant involving two indices. So for your problem above.

x = a[i]+b[j]. Since both arrays are sorted, we can use to pointers by using the first pointer to "watch" the first array a, and using the second pointer to navigate b.

observe that if b[j] < x-a[i] it means we should move forward on the second array.

if b[j] > x-a[i] then for all g > j we can't get the answer so i+=1;

Notice how both indices are moving in accord, they depend on each other.

Hope this helps.

the code, that solves this problem, looks like this:

//n - the length of the array A, m - the length of the array B
int j=m-1;
for (int i=0; i<n; ++i)
{
    while ((j>0) && (A[i]+B[j]>x))
    {
        j--;
    }
    if (A[i]+B[j]==x)
    {
        cout<<i<<" "<<j<<endl;
        return 0;
    }
}
cout<<"Not found!"<<endl;

I din't run the code anywhere, so it may contain some bugs.

The idea is as following. Consider the pair of indexes i1, i2, such as i2>i1. The following assumption is correct: if (A[i1]+B[j1] == x) and (A[i2]+B[j2] == x) then j2<=j1.

So in our code for each i we find the first index j, such as A[i]+B[j]<=x. If (A[i]+B[j]==x), then we find our pair. Otherwise, we continue looking. The total complexity is O(n+m), because i always increases and j always decreases

I prefer using only one loop. I think its more beautiful.

for(int i = 0, j = m - 1; i < n && j >= 0;) {
    if(a[i] + b[j] > x) {
        j--;
    } else if(a[i] + b[j] < x) {
        i++;
    } else {
        printf("%d + %d = %d\n", a[i], b[j], x);
        i++; j--;
    }
}

The idea is as same as the other people's.

Two years later, I am looking for some stuffs on two pointer to send to a friends and I see my name on a comment :)