Proposed resolution:

This wording is relative to N4964.

1. Modify [range.drop.view] as indicated:

   constexpr auto begin()
     requires (!(simple-view<V> &&
                 random_access_range<const V> && sized_range<const V>));
   constexpr auto begin() const
     requires random_access_range<const V> && sized_range<const V>;
   

   -3- Returns:

   1. (?.?) — If V models random_access_range and sized_range,

      ranges::begin(base_) + (ranges::distance(base_) - range_difference_t<V>(size()))
      

   2. (?.?) — Otherwise, ranges::next(ranges::begin(base_), count_, ranges::end(base_)).

   -4- Remarks: In order to provide the amortized constant-time complexity required by the range concept when Vdrop_view does not models random_access_range and sized_rangeforward_range, this functionthe first overload caches the result within the drop_view for use on subsequent calls.

   [Note 1: Without this, applying a reverse_view over a drop_view would have quadratic iteration complexity. — end note]

   
   constexpr auto begin() const
     requires random_access_range<const V> && sized_range<const V>;
   

   -?- Returns: ranges::begin(base_) + (ranges::distance(base_) - range_difference_t<const V>(size())).

[ 2025-10-22; Reflector poll. Status changed: New → Tentatively NAD ]

Set priority to 3 after reflector poll, status to Tentatively NAD.

"NAD, it's Returns: not 'Effects: Equivalent to ...', implementations need to implement it to return that while meeting the amortized 𝒪(1) guarantee."

drop_view::begin const is specified to return ranges::next(ranges::begin(base_), count_, ranges::end(base_)), which has 𝒪(n) complexity when base_ is a random-access-sized but non-common range (demo):

#include <ranges>

int main() {
  const auto s = std::ranges::subrange(std::views::iota(0uz), size_t(-1));
  const auto r = std::ranges::drop_view(s, s.size() - 1);
  const auto b = r.begin(); // time out
}