## Challenge Statement

- Given a
**1-indexed**array of integers*numbers*that is already, find two numbers such that they add up to a specific target*sorted in non-decreasing order**number*. Let these two numbers be*numbers[index1]*and*numbers[index2]*where*1 <= index1 < index2 <= len(numbers)*. - Return
*the indices of the two numbers, index1 and index2,**added by one**as an integer array [index1, index2] of length 2.* - The tests are generated such that there is
**exactly one solution**. You**may not**use the same element twice. - Your solution must use only
**constant extra space**. - This challenge corresponds to LeetCode #167.

### Constraints

`2 <= len(numbers) <= 3 * 10`

^{4}`-1000 <= numbers[i] <= 1000`

*numbers*is sorted in**non-decreasing order**.`-1000 <= target <= 1000`

- The tests are generated such that there is
**exactly one solution.** *s*consists only of printable ASCII characters.

**Example 1:**

**Input:** `numbers = [2, 7, 11, 15]`

, `target = 9`

**Output:** `[1, 2]`

**Explanation:** The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].

**Example 2:**

**Input:** `numbers = [2, 3, 4]`

, `target = 6`

**Output:** `[1, 3]`

**Explanation:** The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].

**Example 3:**

**Input:**` numbers = [-1, 0]`

, `target = -1`

**Output:** `[1, 2]`

**Explanation:** The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].

## Solution

Below is my solution and some test cases. This solution has a **linear time complexity O(n) and a constant space complexity O(1)**, where n is the length of the input list*.*