How does the erase-remove idiom work with ranges/constrained algorithms?

Another option would be decomposing the subrange returned by std::ranges::remove/unique, and use those iterators:

auto [Beg, End] = std::ranges::remove(v, 42);
v.erase(Beg, End);

It doesn't work since std::ranges::remove() returns not iterator but range. But even if you try v.erase(std::ranges::remove(...)) it will not work, because vector does not have erase() overload which takes range as parameter.

Instead, take a look at std::erase() (defined in <vector>). What you need is probably just std::erase(v, 42).

std::ranges::unique (and std::ranges::remove) returns a sub range from the first removed element to the end of the container so you need to use std::begin before passing to std::vector::erase:

v.erase(std::ranges::begin(std::ranges::remove(v, 42)), std::end(v));
v.erase(std::ranges::begin(std::ranges::unique(v)), std::end(v));