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Two Sum II LC #167 Medium Two Pointers
Problem

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. Return their indices as [index1, index2] (1-indexed). You may not use the same element twice. Use only constant extra space.

Example 1
Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: numbers[1] + numbers[2] = 2 + 7 = 9.
Example 2
Input: numbers = [2,3,4], target = 6
Output: [1,3]
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.
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Approach
Two Pointer on Sorted Array
L at start, R at end. Sum too small → L++ (get bigger left value). Sum too big → R-- (get smaller right value). Sum = target → found! O(n) time, O(1) space.
Array State
sum = ?
Variables
L
—
R
—
Sum
—
Target
—
Step Logic
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Algorithm
1
L=0, R=n-1
2
sum = nums[L] + nums[R]
3
sum == target → return [L+1, R+1]
4
sum < target → L++  |  sum > target → R--
Time
O(n)
Space
O(1)
Why It Works

Array is sorted. Sum too small → only moving L right can increase it (left value grows). Sum too large → only moving R left can decrease it (right value shrinks). This monotonicity guarantees we find the unique solution in a single pass — no brute-force needed.