Leetcode 167. Two Sum II – Input Array Is Sorted

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.

Since the numbers are ordered by asc, then we can know the numbers are from smaller to bigger.

Using two pointers to match the target. If the sum is smaller, then it means we need to move left pointer to get bigger number. If the sum is bigger, then it means we need to move right pointer to get smaller number.

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        l = 0
        r = len(numbers) - 1

        while numbers[l] + numbers[r] != target:
            if numbers[l] + numbers[r] > target:
                r -= 1
            else:
                l += 1
        
        return [l+1, r+1]