Question

Leetcode link -> leetcode.com/problems/two-sum-ii-input-arra..

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].

Approach

1. Initialize two pointers:
2. left starts at the first element (left = 0).

3. right starts at the last element (right = len(numbers) - 1).

4. While loop:

5. As long as left is less than right, calculate the sum of numbers[left] and numbers[right].
6. If the sum equals the target, return the 1-based indices [left + 1, right + 1].
7. If the sum is less than the target, increment left to increase the sum.
8. If the sum is greater than the target, decrement right to decrease the sum.

   Complexity

Time

𝑂(𝑛)

Space

𝑂(1)

Code

from typing import List
from collections import defaultdict


class Solution:

    def two_sum(self, nums:List[int], target:int) -> List[int]:
        left = 0
        right = len(nums) - 1

        while left < right:
            current_sum = nums[left] + nums[right]

            if current_sum < target:
                left += 1
            elif current_sum > target:
                right -= 1
            else:
                return [left + 1, right + 1]