Pick one

• 1Choice 1.

  With negatives, extending the right end can shrink the sum, so the monotonicity behind "shrink from the left when the sum exceeds the limit" breaks and two pointers can miss the optimum. You need prefix sums combined with a monotonic queue or binary search, or another approach

• 2Choice 2.

  Insert negatives as their absolute values and the same algorithm works unchanged

• 3Choice 3.

  Reset the window the moment the sum goes negative and it stays O(n)

• 4Choice 4.

  No problem. Two pointers always give the right answer in O(n) regardless of element sign