Two Sum II - Input array is sorted

Problem Statement - link #

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 <= first < second <= numbers.length.

Return the indices of the two numbers, index1 and index2, 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.

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.

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.

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.

Constraints #

• 2 <= numbers.length <= 3 * 10^4
• -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:
    int find2ndIndex(vector<int>& numbers,int left,int right,int t){
        while(left<=right){
            int mid = left + (right-left)/2;
            if(numbers[mid] == t){
                return mid;
            }else if(numbers[mid] > t){
                right = mid-1;
            }else{
                left = mid+1;
            }
        }
        return 0;
    }
    vector<int> twoSum(vector<int>& numbers, int target) {     
        for(int i=0; i<numbers.size();i++){
            int t = target - numbers[i];
            int left=i+1,right=numbers.size()-1,mid; 
            mid = find2ndIndex(numbers,left,right,t);
            if(mid)
                return {i+1,mid+1};
        }
        return {0,0};
    }
};