two-sum-ii py3
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0167_two-sum-ii-input-array-is-sorted/README.md
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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`.
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Return _the indices of the two numbers,_ `index1` _and_ `index2`_, **added by one** as an integer array_ `[index1, index2]` _of length 2._
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The tests are generated such that there is **exactly one solution**. You **may not** use the same element twice.
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Your solution must use only constant extra space.
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**Example 1:**
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Input: numbers = [2,7,11,15], target = 9
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Output: [1,2]
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Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].
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**Example 2:**
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Input: numbers = [2,3,4], target = 6
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Output: [1,3]
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Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].
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**Example 3:**
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Input: numbers = [-1,0], target = -1
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Output: [1,2]
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Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].
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**Constraints:**
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* `2 <= numbers.length <= 3 * 104`
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* `-1000 <= numbers[i] <= 1000`
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* `numbers` is sorted in **non-decreasing order**.
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* `-1000 <= target <= 1000`
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* The tests are generated such that there is **exactly one solution**.
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0167_two-sum-ii-input-array-is-sorted/python3/solution.py
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0167_two-sum-ii-input-array-is-sorted/python3/solution.py
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class Solution:
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def twoSum(self, numbers: List[int], target: int) -> List[int]:
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left = 0
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right = len(numbers) - 1
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while left < right:
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total = numbers[left] + numbers[right]
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if total == target:
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return [left + 1, right + 1]
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if total > target:
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right -= 1
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else:
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left += 1
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