Two Sum II - Input Array Is Sorted

In this task, I worked on finding two numbers in a sorted array that add up to a given target value. This problem helped me understand how to efficiently use pointers instead of checking all possible pairs.

What I Did

I created a function twoSum that takes a sorted list of numbers and a target value. The function returns the positions of the two numbers whose sum equals the target.

For example:
Input: numbers = [2, 7, 11, 15], target = 9
Output: [1, 2]

This means the numbers at positions 1 and 2 add up to 9.

How I Solved It

To solve this, I used two pointers:

• l starting from the beginning of the array
• r starting from the end of the array

Then I checked the sum of the elements at these positions.

At each step:

• If the sum equals the target, I return their positions
• If the sum is less than the target, I move the left pointer forward
• If the sum is greater than the target, I move the right pointer backward

Code

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

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

How It Works

Since the array is already sorted, this approach works efficiently by narrowing down the search space from both ends.

Instead of checking every pair, the two-pointer method quickly finds the correct pair in a single pass.