# Two Sum II - Input array is sorted

## Description

Given an array of integers that is already _**sorted in ascending order**_, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.

**Note:**

* Your returned answers (both index1 and index2) are not zero-based.
* You may assume that each input would have _exactly_ one solution and you may not use the _same_ element twice.

**Example:**

```
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.
```

## Solution

```cpp
class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int left = 0, right = numbers.size() - 1;
        while(left < right){
            int sum = numbers[left] + numbers[right];
            if(sum < target)
                ++left;
            else if(sum > target)
                --right;
            else
                break;
        }
        return {left + 1, right + 1};
    }
};
```

### Idea of proof:

Loop invariant: it is not possible for any number in `nums[0], ..., nums[left-1], nums[right+1], ..., nums[n-1]` to be one of the two numbers that sum up to `target`.
