Leetcode(Medium) - Two Sum II - Input Array Is Sorted

Question

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 <= nums.length <= 3 * 10^4
• -1000 <= nums[i] <= 1000
• numbers is sorted in non-decreasing order.
• -1000 <= target <= 1000
• The tests are generated such that there is exactly one solution.

Solution

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        left, right = 0, len(numbers)-1
        increment = lambda x: x + 1
        while left < right:
            targetSum = numbers[left] + numbers[right]
            if targetSum < target:
                left += 1
            elif targetSum > target:
                right -= 1
            else:
                return list(map(increment, [left, right]))

Time/Space Complexity

• Time Complexity: O(n)
• Space Complexity: O(1)