For all a_i, Find max j where a_j is (> a_i and <= a_i + d) and( < a_i and >= a_i — d ).

You can use binary search for this

We can scan through given array from right to left, supporting set of indices, answer for which has not been found yet.
Let's suppose we are now in index $$$j$$$ and we want to find all $$$i$$$ s.t. $$$j$$$ is the answer for $$$i$$$. These $$$i$$$'s values are in range $$$[v[j] - d, v[j] + d]$$$, so just finding, processing and deleting all such $$$i$$$s suffices. As I understood, the answer for particular $$$i$$$ can never be $$$-1$$$, because $$$a[i] - a[i] = 0$$$.

void RunCase([[maybe_unused]] int testcase) {
  int n;
  int d;
  std::cin >> n >> d;

  std::vector<int> v(n);
  for (int& i : v) {
    std::cin >> i;
  }

  // (value, index)
  std::set<std::pair<int, int>> active;
  for (int i = 0; i < n; ++i) {
    active.emplace(v[i], i);
  }

  std::vector<int> ans(n);
  std::iota(ans.begin(), ans.end(), 0);
  for (int i = n - 1; i >= 0; --i) {
    auto it = active.lower_bound(std::pair(v[i] - d, -1));
    while (it != active.end() && it->first <= v[i] + d) {
      ans[it->second] = i;
      it = active.erase(it);
    }
  }

  for (int i = 0; i < n; ++i) {
    std::cout << ans[i] + 1 << " \n"[i == n - 1];
  }
}