Skip to content

167. Two Sum II - Input Array Is Sorted

Description

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.

Solutions

Note that the array is sorted in non-decreasing order, so for each numbers[i], we can find the position of target - numbers[i] by binary search, and return $[i + 1, j + 1]$ if it exists.

The time complexity is $O(n \times \log n)$, where $n$ is the length of the array numbers. The space complexity is $O(1)$.

1
2
3
4
5
6
7
8
class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        n = len(numbers)
        for i in range(n - 1):
            x = target - numbers[i]
            j = bisect_left(numbers, x, lo=i + 1)
            if j < n and numbers[j] == x:
                return [i + 1, j + 1]
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
class Solution {
    public int[] twoSum(int[] numbers, int target) {
        for (int i = 0, n = numbers.length;; ++i) {
            int x = target - numbers[i];
            int l = i + 1, r = n - 1;
            while (l < r) {
                int mid = (l + r) >> 1;
                if (numbers[mid] >= x) {
                    r = mid;
                } else {
                    l = mid + 1;
                }
            }
            if (numbers[l] == x) {
                return new int[] {i + 1, l + 1};
            }
        }
    }
}
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        for (int i = 0, n = numbers.size();; ++i) {
            int x = target - numbers[i];
            int j = lower_bound(numbers.begin() + i + 1, numbers.end(), x) - numbers.begin();
            if (j < n && numbers[j] == x) {
                return {i + 1, j + 1};
            }
        }
    }
};
1
2
3
4
5
6
7
8
9
func twoSum(numbers []int, target int) []int {
    for i, n := 0, len(numbers); ; i++ {
        x := target - numbers[i]
        j := sort.SearchInts(numbers[i+1:], x) + i + 1
        if j < n && numbers[j] == x {
            return []int{i + 1, j + 1}
        }
    }
}
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
function twoSum(numbers: number[], target: number): number[] {
    const n = numbers.length;
    for (let i = 0; ; ++i) {
        const x = target - numbers[i];
        let l = i + 1;
        let r = n - 1;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (numbers[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        if (numbers[