# 0167-two-sum-ii-input-array-is-sorted Try it on leetcode ## 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, each incremented 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:

## Solution(Python) ```Python class Solution: def twoSum(self, numbers: List[int], target: int) -> List[int]: # numbers = [2,7,11,15], target = 9 # i = 0 j = 3 ; 2 + 15 = 17 > 9 f # j =2 # i=0 ; j =2; 2 + 11 =13 > 9 f # j =1 # i = 0; j =1; 2 + 7 = 9; 9 == 9 t # # numbers = [2,3,4], target = 6 # i =0 ; i =2 6 == 6 t # # i = 0 j =n -1 # invariant i < j: # i + j > target reduce j # else: # increase i # if match return i j # edge cases # i = 0 j = len(numbers) - 1 while i < j: cursum = numbers[i] + numbers[j] if cursum == target: return [i+1, j +1] elif cursum > target: j -= 1 else: i +=1 ```