No. Taking 5 excludes both 4s. Taking the two endpoints gives 8. The future effects of a local choice matter, so compare the complete consequences of taking and skipping.

Let best(i) be the maximum total from positions i onward, with no earlier choice restricting position i. Picking nothing is allowed.

If you skip, the answer is best(i+1). If you take, it is values[i] + best(i+2). Every valid choice falls into one of these cases, so:

best(i) = max(best(i+1), values[i] + best(i+2)).

from functools import cache

def best_non_adjacent(values):
    @cache
    def best(i):
        if i >= len(values):
            return 0
        return max(best(i + 1), values[i] + best(i + 2))
    return best(0)

assert best_non_adjacent([4, 5, 4]) == 8
assert best_non_adjacent([-5, -1]) == 0
assert best_non_adjacent([]) == 0

The single independent coordinate is i. There are O(n) states, each with two choices, yielding O(n) time and O(n) cache/stack space. A reverse-index table evaluates the same recurrence.

Why no previous_was_taken flag? Taking position i jumps directly to i+2, where no earlier selection restricts the current position. The contract and transition have already encoded that restriction. A different formulation that advances one step every time may need a flag.

Analogy limit: “1D” does not mean “easy,” nor does it mean there is only one valid formulation. It describes the chosen state, not a property stamped permanently on the problem.

No. For length at least two, solve two linear cases: exclude the first position, or exclude the last, and take the better result. Handle the empty and single-position inputs separately. Every valid circular choice excludes at least one endpoint.