Two Sum (Sorted Array)

In this task, I worked on finding two numbers in a sorted array that add up to a given target.
The goal is to return their 1-based indices.

What I Did
I implemented a function twoSum that takes:
a sorted array numbers
a target value
and returns the indices of the two numbers whose sum equals the target.

Example:
Input: numbers = [2, 7, 11, 15], target = 9

Output: [1, 2]

Explanation:
2 + 7 = 9 → indices are 1 and 2
How I Solved It

Instead of brute force (O(n²)), I used the Two Pointer Technique.

Core Idea:

Since the array is sorted:

• If sum is too small → move left pointer forward
• If sum is too large → move right pointer backward
• Step-by-step approach:
• Initialize two pointers:
• left = 0
• right = len(numbers) - 1
• Loop while left < right:
• Calculate sum = numbers[left] + numbers[right]
• If sum equals target → return indices
• If sum < target → move left forward
• If sum > target → move right backward CODE:

class Solution:
    def twoSum(self, numbers, target):
        left = 0
        right = len(numbers) - 1

        while left < right:
            total = numbers[left] + numbers[right]

            if total == target:
                return [left + 1, right + 1]  # 1-based index

            elif total < target:
                left += 1
            else:
                right -= 1 

How It Works
The algorithm narrows down the search space using two pointers
It avoids checking all pairs
Uses the sorted property to make smart moves