167. Two Sum II - Input Array Is Sorted

167. Two Sum II - Input Array Is Sorted LeetCode solution

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.

Solution:

class Solution {

public:

    vector<int> twoSum(vector<int>& numbers, int target) {

        vector<int>vt;

        int n=numbers.size();

        int l=0, r=n-1;

        while(l<r)

        {

            if(numbers[l]+numbers[r]==target)

            {

                vt.push_back(l+1);

                vt.push_back(r+1);

                break;

            }

            else if(numbers[l]+numbers[r]<target)

            {

                l++;

            }

            else if(numbers[l]+numbers[r]>target)

            {

                r--;

            }

        }

        return vt;

    }

};