notes/OJ notes/pages/Leetcode Two-Sum-II-Input-Array-Is-Sorted.md

# Leetcode Two-Sum-II-Input-Array-Is-Sorted

#### 2022-07-11 14:54

> ##### Algorithms:
> #algorithm #two_pointers 
> ##### Data structures:
> #DS #array 
> ##### Difficulty:
> #coding_problem #difficulty-easy 
> ##### Additional tags:
> #leetcode 
> ##### Revisions:
> N/A

##### Related topics:
```expander
tag:#two_pointers
```
 
- [[Leetcode Intersection-of-Two-Arrays-II]]
- [[Leetcode Merge-Sorted-Array]]
- [[Leetcode Merge-Two-Sorted-Lists]]
- [[Leetcode Move-Zeroes]]
- [[Leetcode Squares-of-a-Sorted-Array]]
- [[Two pointers approach]]
 

##### Links:
- [Link to problem](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/)
___
### Problem

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.

#### Examples

**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**.

### Thoughts

> [!summary]
> This is a #two_pointers problem.

decrement the right pointer makes the sum smaller, increment the left pointer makes the sum bigger, and thud we get the answer.

### Solution

```cpp
class Solution {
public:
  vector<int> twoSum(vector<int> &numbers, int target) {
    // Sorted array, use two pointers
    // remember to subtract 1
    int r = numbers.size() - 1;
    int l = 0;
    int sum;

    while (l < r) {
      sum = numbers[l] + numbers[r];
      if (sum < target) {
        l++;
      } else if (sum > target) {
        r--;
      } else {
        return {l + 1, r + 1};
      }
    }

    // Shouldn't return this, since there is a solution
    return {0, 0};
  }
};
```