167

167. Two Sum II - Input Array Is Sorted (Medium)

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.

Companies: Amazon, Adobe, Bloomberg

Related Topics:
Array, Two Pointers, Binary Search

Similar Questions:

• Two Sum (Easy)
• Two Sum IV - Input is a BST (Easy)
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Solution 1. Two Pointers

// OJ: https://leetcode.com/problems/two-sum-ii-input-array-is-sorted
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> twoSum(vector<int>& A, int target) {
        int i = 0, j = A.size() - 1;
        while (i < j) {
            if (A[i] + A[j] == target) return {i + 1, j + 1};
            if (A[i] + A[j] < target) ++i;
            else --j;
        }
        return {};
    }
};