LC 167 - 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, $\text{index}_1$ and  $\text{index}_2$, added by one as an integer array $[\text{index}_1, \text{index}_2]$ 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, $\text{index}_1$ = 1, $\text{index}_2$ = 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 $\text{index}_1$ = 1, $\text{index}_2$ = 3. We return [1, 3].

Example 3:

Input: numbers = [-1,0]`, target = -
Output: [1,2]

Explanation: The sum of -1 and 0 is -1. Therefore $\text{index}_1$ = 1, $\text{index}_2$ = 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.

Links

Question here and Solution here

Solution

concept

Since the array is sorted, we use a left and right pointer to keep track the curSum = numbers[l] + numbers[r]. If curSum > target, we will move right to the left and if curSum < target, we will move left to the right. If the condition curSum == target, then it is the answer. This works because the array is sorted.

code

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

        while l < r:
            curSum = numbers[l] + numbers[r]

            if curSum > target:
                r -= 1
            elif curSum < target:
                l += 1
            else:
                return [l + 1, r + 1]
        return []

Complexity

time: $O(n)$
space: $O(1)$