Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index<sub>1</sub>] and numbers[index<sub>2</sub>] where 1 <= index<sub>1</sub> < index<sub>2</sub> < numbers.length.

Return the indices of the two numbers, index<sub>1</sub> and index<sub>2</sub>, added by one as an integer array [index<sub>1</sub>, index<sub>2</sub>] 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.

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].

Constraints:

• 2 <= numbers.length <= 3 * 10<sup>4</sup>

• -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.

Using two pointers.

If sum is smaller than target then left++

If sum is bigger than target then right--

Until sum is equal to target.

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int left = 0;
        int right = numbers.length - 1;

        while (left < right){
            if (numbers[left] + numbers[right] > target){
                right -= 1;
            } else if (numbers[left] + numbers[right] < target) {
                left += 1;
            } else {
                return new int[]{left + 1, right + 1};
            }    
        }
        return new int[0];
    }
}