167. Two Sum II - Input Array Is Sorted

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167. Two Sum II - Input Array Is Sorted

Description

Difficulty: Medium

Related Topics: Array, Two Pointers, Binary Search

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:

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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:

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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:

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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.

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int left = 0, right = numbers.size() - 1;
        while (left < right) {
            int sum = numbers[left] + numbers[right];
            if (sum == target) {
                return {left + 1, right + 1};
            } else if (sum < target) {
                left++;
            } else if (sum > target) {
                right--;
            }
        }
        return {-1, -1};
    }
};