# [Solved] 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[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.

## Question

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[index1]` and `numbers[index2]` where `1 <= index1 < index2 <= numbers.length`.

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.

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 * 104`
• `-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.

## Python Solution

```class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
d={}
for i in range(len(numbers)):
if numbers[i] in d:
return(d[numbers[i]]+1,i+1)
else:
d[target-numbers[i]] = i```