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Two Sum Pattern

2 + 6 = 8

L=0target: 10R=9

Two Sum II

Use two pointers starting at opposite ends of a sorted array. Converge inward based on whether the current sum is too small or too large — O(n) time, O(1) space.

Visualization

Target:

Speed

Comparisons

Searching

Status

2 + 40 = 42

Current Sum

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Left pointer (L)Right pointer (R)Found pairBetween pointers

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Complexity

O(1)

Best Case

O(n)

Worst Case

O(n)

Average

O(1)

Space

How It Works

1Place left pointer at index 0, right pointer at index n-1
2Compute sum = arr[left] + arr[right]
3If sum == target: return [left+1, right+1]. If sum < target: left++. If sum > target: right--
4Repeat until left >= right (not found) or pair located

Why Sorted Order is Required

•Sorted order gives a decision rule: sum too small means move left right; sum too large means move right left
•Without sorting the same sum could be achieved by many pairs, so no single pointer move is safe
•The problem name "Two Sum II" refers to LeetCode 167 where the input is guaranteed sorted

Active Recall Quiz

Question 1 of 3