# 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

Solution 1: Binary Search

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 $O(1)$.

Solution 2: Two Pointers

We define two pointers $i$ and $j$, which point to the first element and the last element of the array respectively. Each time we calculate $numbers[i] + numbers[j]$. If the sum is equal to the target value, return $[i + 1, j + 1]$ directly. If the sum is less than the target value, move $i$ to the right by one position, and if the sum is greater than the target value, move $j$ to the left by one position.

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

• class Solution {
public int[] twoSum(int[] numbers, int target) {
for (int i = 0, j = numbers.length - 1;;) {
int x = numbers[i] + numbers[j];
if (x == target) {
return new int[] {i + 1, j + 1};
}
if (x < target) {
++i;
} else {
--j;
}
}
}
}

• class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
for (int i = 0, j = numbers.size() - 1;;) {
int x = numbers[i] + numbers[j];
if (x == target) {
return {i + 1, j + 1};
}
if (x < target) {
++i;
} else {
--j;
}
}
}
};

• class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
i, j = 0, len(numbers) - 1
while i < j:
x = numbers[i] + numbers[j]
if x == target:
return [i + 1, j + 1]
if x < target:
i += 1
else:
j -= 1


• func twoSum(numbers []int, target int) []int {
for i, j := 0, len(numbers)-1; ; {
x := numbers[i] + numbers[j]
if x == target {
return []int{i + 1, j + 1}
}
if x < target {
i++
} else {
j--
}
}
}

• function twoSum(numbers: number[], target: number): number[] {
for (let i = 0, j = numbers.length - 1; ; ) {
const x = numbers[i] + numbers[j];
if (x === target) {
return [i + 1, j + 1];
}
if (x < target) {
++i;
} else {
--j;
}
}
}


• /**
* @param {number[]} numbers
* @param {number} target
* @return {number[]}
*/
var twoSum = function (numbers, target) {
for (let i = 0, j = numbers.length - 1; ; ) {
const x = numbers[i] + numbers[j];
if (x === target) {
return [i + 1, j + 1];
}
if (x < target) {
++i;
} else {
--j;
}
}
};


• use std::cmp::Ordering;

impl Solution {
pub fn two_sum(numbers: Vec<i32>, target: i32) -> Vec<i32> {
let n = numbers.len();
let mut l = 0;
let mut r = n - 1;
loop {
match (numbers[l] + numbers[r]).cmp(&target) {
Ordering::Less => {
l += 1;
}
Ordering::Greater => {
r -= 1;
}
Ordering::Equal => {
break;
}
}
}
vec![(l as i32) + 1, (r as i32) + 1]
}
}