167. Two Sum II — Input Array Is Sorted

At a Glance

Topic: two-pointers
Pattern: Two Pointers
Difficulty: Medium
Companies: Amazon, Google, Meta, Apple, Adobe
Frequency: High
LeetCode: 167

Given a 1-indexed sorted strictly increasing array numbers and target, return the 1-indexed pair [leftIndex, rightIndex] with leftIndex < rightIndex such that numbers[leftIndex] + numbers[rightIndex] == target. Exactly one solution exists.

Recognition Cues

Sorted array + pair sum
"Two pointers" classic
1-indexed output (LeetCode quirk)

Diagram

At-a-glance flow (replace with problem-specific Mermaid as you refine this note). camelCase node IDs; no spaces in IDs.

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Approaches

Brute force — O(n²) pairs.
Better — binary search for complement — O(n log n).
Optimal — O(n) time / O(1) space — inward pointers: too small → move left up; too large → move right down.

Optimal Solution

Go

func twoSum(numbers []int, target int) []int {
	left, right := 0, len(numbers)-1
	for left < right {
		sum := numbers[left] + numbers[right]
		if sum == target {
			return []int{left + 1, right + 1}
		}
		if sum < target {
			left++
		} else {
			right--
		}
	}
	return nil
}

JavaScript

function twoSum(numbers, target) {
	let left = 0;
	let right = numbers.length - 1;
	while (left < right) {
		const sum = numbers[left] + numbers[right];
		if (sum === target) {
			return [left + 1, right + 1];
		}
		if (sum < target) {
			left++;
		} else {
			right--;
		}
	}
	return [];
}