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## Thoughts

Easy Problem. The solution will only work when the array is already sorted

## Complexity Analysis

**Time Complexity: O(n)**

The algorithm uses two pointers starting at opposite ends of the array, moving toward each other. In the worst case (no pair found), every element is visited exactly once before `start` and `end` meet. Since each pointer only ever moves inward and never backtracks, you get at most `n` iterations total.

**Space Complexity: O(1)**

Only two integer variables (`start` and `end`) are used regardless of input size. The returned slice is always a fixed size of 2 elements, so it doesn't scale with input and isn't counted as auxiliary space.

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**Why this is efficient:** The naive brute-force approach of checking every pair would be O(n²) time. This algorithm achieves linear time by exploiting the fact that the array is sorted — if the current sum is too large, you know moving `end` left is the only useful action, and vice versa for `start`. This eliminates the need for a nested loop entirely.

First solved

Mar 2, 2026