Two Sum II - Input Array Is Sorted

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.

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

array is sorted 
Should not use the extra space O(1)
two pointers, one starting from the left and start from the right
if sum = left + right == target return both index 
if sum = left + right < target right--
if sum = left + right > target left++

return the empty integer array

Follow Up:

Sum = left + right // In that case, to prevent an overflow error, we may define the sum as the long

Complexity:

Time complexity: O(n).
The input array is traversed at most once. Thus the time complexity is O(n).

Space complexity: O(1).
We only use additional space to store two indices and the sum, so the space complexity is O(1).

Code


class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int left = 0, right = numbers.length -1;

        while(left <= right){
            long sum = numbers[left] + numbers[right];
            if(sum == target){
                return new int[]{left+1, right+1};
            }
            if (sum > target){
                right--;
            }
            if(sum < target){
                left++;
            }
        }

        return new int[]{};
    }
}

Happy Coding
Vishwaraj Sali
sali.vishwaraj@gmail.com
https://www.Vishsali.dev