关于C++中std::includes算法时间复杂度的疑问

Great question! Let's unpack why the standard and references state that std::includes makes at most 2*(N1 + N2 - 1) comparisons, even though your reasoning about the worst case with max(set2) >= max(set1) seems to suggest a tighter bound of 2*N1.

First, Let's Recap How std::includes Works #

std::includes operates on two sorted ranges (using the same ordering, e.g., ascending). It uses a two-pointer approach:

• Initialize it1 at the start of the first range (set1, size N1) and it2 at the start of the second range (set2, size N2).
• In each iteration:
  1. If *it1 < *it2: Move it1 forward (1 comparison).
  2. Else if *it2 < *it1: Return false immediately (2 comparisons total for this step).
  3. Else (elements are equal): Move both pointers forward (2 comparisons total for this step).
• The loop ends when either pointer reaches the end of its range. If it2 has reached the end, return true; otherwise, return false.

Why the Standard Uses 2*(N1 + N2 - 1) as the Upper Bound #

Your observation about 2*N1 being a bound for cases where max(set2) >= max(set1) is correct for that specific scenario—once it1 exhausts set1, we know set2 can't be fully contained, so we stop. But the standard's upper bound is a conservative, universal guarantee that covers all possible input cases, not just this one.

Here's why the larger bound exists:

1. Worst-Case Iterations with Two Comparisons Per Step: In scenarios where we need to traverse nearly all elements of both ranges and every iteration requires two comparisons, the total count can approach 2*(N1 + N2 - 1). For example:

   • Imagine set1 is [1, 2, 3, ..., N1] and set2 is [1, 2, 3, ..., N2] where N2 = N1. Every iteration requires two comparisons (to confirm elements are equal) before moving both pointers. That's 2*N2 comparisons, which is less than 2*(N1 + N2 -1) but shows how two comparisons per step add up.
   • For cases where elements alternate between the two ranges but still require full traversal (e.g., set1 = [1,3,5,...,N1] and set2 = [1,3,5,...,N2] with N2 < N1), we end up with a mix of one and two comparison steps, but the total never exceeds the standard's bound.

2. A Universal Upper Bound: The standard's goal is to provide a single upper limit that holds for all valid inputs, regardless of how the elements are arranged. While your 2*N1 bound is tight for one specific worst case, 2*(N1 + N2 -1) ensures that no matter what sorted ranges you pass in, the algorithm will never exceed this number of comparisons. It's a safe, all-encompassing upper limit derived from the algorithm's structure (linear traversal with up to two comparisons per step).

In Short #

Your reasoning about the max(set2) >= max(set1) scenario is correct for that case, but the standard's upper bound is a broader guarantee that accounts for all possible sorted input combinations. It's a conservative estimate that ensures the algorithm's performance stays within predictable linear bounds.

内容的提问来源于stack exchange，提问作者MrPisarik