Prompt
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, each incremented 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.
Constraints:
2 <= numbers.length <= 3 * 104-1000 <= numbers[i] <= 1000numbersis sorted in non-decreasing order.-1000 <= target <= 1000- The tests are generated such that there is exactly one solution.
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].
Solution
In C++
vector<int> twoSum(vector<int>& nums, int target) {
int l = 0; int r = nums.size() - 1;
while (l < r) {
int sum = nums[l] + nums[r];
if (sum < target) l++;
else if (sum > target) r--;
else return {l + 1, r + 1};
}
return {};
}Explanation
Because the array is sorted, move the pointers in as needed until we hit the target.
Big O Analysis
Time Complexity