Apartments Problem

#1

CSES problem link: Apartments

I first sorted both the applicant and apartment arrays. For applicant x, I want to select the least apartment that is within the range [x-k, x+k]. I take the smallest apartment i because if I choose j > i, there will be at most the same number of ways to give apartments to the remaining applicants. Since applicant y > x has y-k > x-k, j could overlap with the range [y-x, y+k] and limit the possibilities for applicant y. So the smallest apartment within the range of x is optimal.

I implement this using two pointers, as follows.

int main()
{
    int n, m, k;
    cin >> n >> m >> k;
    vi a(n), b(m);
    F0R(i, n) cin >> a[i];
    F0R(i, m) cin >> b[i];
    sort(all(a)); //applicants
    sort(all(b)); //apartments
    int count = 0, j = 0;
    trav(i, a) {
      // starting from j, what is the first apartment that works?
      while (j < m && b[j] < i-k)
          j++;
      if (j == m) break;
      if (j <= i+k) {
        count++;
        j++; // apartment is sold, look for a new one
      }
    }
    cout << count << endl;
}

This gives a wrong answer on many test cases.

Since I am weak at greedy algorithms, I want to be able to think of greedy methods myself. Please do not reveal the approach to this problem, but please explain why my idea is wrong and how I might start over. To be clear, I would appreciate a nudge in the right direction, not a full thought process.
Thanks!

#2

Your idea is correct but this line should be
if(b[j] <= i+k)

#3

Thank you! What a silly mistake. I learned that I need to be more careful and write "index j and apartment b[j]" if necessary. While debugging, I will also look over my code carefully to spot errors such as replacing an index with the array value of that index. Thanks for helping me with this.