Two Sum II — Sorted

Problem

Given a 1-indexed sorted array nums (ascending) and a target, return the 1-indexed positions of the two numbers that sum to target. There is exactly one solution.

Constraints

• 2 ≤ nums.length ≤ 3·10⁴
• −1000 ≤ nums[i] ≤ 1000
• sorted ascending

Approach — Two Pointers

This is a Two Pointers problem. The idea: walk two indices through the data together (or toward each other), turning an O(n²) pair search into one linear pass. Work through the reference code below line by line, then re-derive it yourself in the editor — that's how the pattern sticks.

Complexity: O(n) time, O(1) extra space.

Solution code

Java

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
        while (l < r) {
            int s = nums[l] + nums[r];
            if (s == target) return new int[]{ l + 1, r + 1 };
            if (s < target) l++;
            else            r--;
        }
        return new int[]{};
    }
}

Practice it

Reading a solution isn't the same as being able to write it under pressure. Open this problem in the in-browser editor, hide the solution, and solve it from scratch — your code runs against real test cases instantly.