No. Each state might scan 1,000 candidates, giving roughly a million candidate checks. State count alone omits transition work. Sorting, copying, and producing the output may add more work.

For a finite DP, a useful estimate is:

time = sum of work over evaluated states

When each state has the same bound, use number of states × work per state. An array recurrence with n states and two options per state is often O(n). A recurrence comparing every earlier position has O(n²) work even if it uses only one DP array.

def predecessor_checks(n):
    return sum(i for i in range(n))

assert predecessor_checks(5) == 10
assert predecessor_checks(1000) == 499500

Typical shapes to recognize: a scan is linear; repeatedly halving a search range takes logarithmically many steps; considering every pair is quadratic; choosing include/exclude for each of n items produces 2**n combinations.

Do not equate nested loops with quadratic time without counting: two pointers that only move forward can each move at most n times in total.

Output matters. Listing all subsets of n distinct items requires 2**n outputs. Copying the elements into separate lists costs Θ(n·2**n) overall. A clever cache cannot make that full output polynomial in n.

Space includes recursion frames and stored output, not just explicit arrays. Big-O here counts abstract operations; Python integers can grow, so arithmetic on enormous counts is not literally constant time.

Analogy limit: estimates are not wall-clock benchmarks. Implementation, constants, input distribution, and available memory still matter.

No. The answer is one integer, 2**n; you no longer have to construct every subset. Its representation still has n + 1 binary digits. Changing the required output can change the whole algorithm.