Two pointers: pair sum in sorted array

The two pointers technique sweeps two indices through a sequence at once, usually from opposite ends, to avoid a nested loop. It shines on sorted data, where the order tells you which pointer to move:

left, right = 0, len(nums) - 1
while left < right:
    s = nums[left] + nums[right]
    # too small? move left up. too big? move right down.

Because nums is sorted, raising left can only increase the sum and lowering right can only decrease it — so each step provably moves you toward the target. This is the classic "Two Sum II" (LeetCode #167) idea.

Your task

Write pair_sum(nums, target) where nums is sorted ascending. Return the two 0-based indices [i, j] (with i < j) whose values add to target. Exactly one solution exists.

pair_sum([2, 7, 11, 15], 9)    # [0, 1]   (2 + 7)
pair_sum([1, 2, 3, 4, 6], 10)  # [3, 4]   (4 + 6)
pair_sum([2, 3, 4], 6)         # [0, 2]   (2 + 4)

Hint: start left at the front and right at the back. If the sum is too small move left up; if too big move right down; if exact, return them.