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Two Sum

Given a sorted array, return the 1-indexed positions of the two numbers that add up to the target.

EASY

Problem intuition

The sorted order is the whole unlock here
If the current sum is too small, only the left pointer can help.
If it is too large, only the right pointer can help
That monotonic movement makes the two-pointer solution both correct and optimal after you move past the brute-force baseline.

Solution

The solutions below are ordered from least optimal to most optimal, so you can see the improvement path instead of only the final answer.

Solution 1

Brute force pair scan

Check every pair until you find the matching sum.
This is simple, but it ignores the fact that the array is already sorted.
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Time complexity: O(n^2)
- the nested loops try every possible pair before returning the match.
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Space complexity: O(1)
- only loop counters and the returned pair are used as extra storage.

Solution 2

Two pointers

Use the sorted property directly instead of scanning every pair.
Move the left pointer when the sum is too small and the right pointer when the sum is too large.
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Time complexity: O(n)
- each pointer moves inward across the sorted array at most once.
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Space complexity: O(1)
- the algorithm keeps only two pointers and a temporary sum.