---
title: 'Two Pointers'
description: 'Converging/parallel index walk that prunes the O(n^2) pair space using order.'
---

## The idea

Checking every pair costs O(n²). That is n² boxes to open.

But if the array is **ordered**, you do not need every pair.
Put one pointer at each end. Look at the sum:
- Too small? Only a bigger left value can help. Move left in.
- Too big? Only a smaller right value can help. Move right in.

Each move kills a whole row of the pair space. O(n²) → O(n).

## The picture

```mermaid
flowchart TD
    A["2  7  11  15   target 18<br/>L           R"] --> B["2 + 15 = 17 < 18<br/>too small → move L right"]
    B --> C["2  7  11  15<br/>    L       R"]
    C --> D["7 + 15 = 22 > 18<br/>too big → move R left"]
    D --> E["2  7  11  15<br/>    L   R"]
    E --> F["7 + 11 = 18 ✓"]

    style F fill:#2e7d32,color:#fff
```

## Not the same as binary search

Binary search **jumps** to the middle and discards half unseen — O(log n).
Two pointers **walks** — every element gets inspected once — O(n).
Both need order. Different moves.

## Two Sum II (LC 167) — the pure template

```python
def two_sum(numbers: list[int], target: int) -> list[int]:
    lo, hi = 0, len(numbers) - 1

    while lo < hi:
        total = numbers[lo] + numbers[hi]
        if total == target:
            return [lo + 1, hi + 1]
        if total < target:
            lo += 1        # only a bigger value helps
        else:
            hi -= 1        # only a smaller value helps
```

## Container With Most Water (LC 11) — move the shorter wall

Water = width × shorter wall.
Moving the taller wall can never help — width shrinks, height cannot grow.
So always move the shorter one. Say this proof in the interview.

```python
def max_area(height: list[int]) -> int:
    lo, hi = 0, len(height) - 1
    best = 0

    while lo < hi:
        best = max(best, (hi - lo) * min(height[lo], height[hi]))
        if height[lo] < height[hi]:
            lo += 1
        else:
            hi -= 1
    return best
```

## 3Sum (LC 15) — fix one, converge two

Sort first. Fix the smallest number. Now it is Two Sum II on the rest.
Skip duplicates at every level or you get repeat triplets.

```python
def three_sum(nums: list[int]) -> list[list[int]]:
    nums.sort()
    result = []

    for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
            continue                          # skip duplicate anchor
        lo, hi = i + 1, len(nums) - 1
        while lo < hi:
            total = nums[i] + nums[lo] + nums[hi]
            if total < 0:
                lo += 1
            elif total > 0:
                hi -= 1
            else:
                result.append([nums[i], nums[lo], nums[hi]])
                lo += 1
                while lo < hi and nums[lo] == nums[lo - 1]:
                    lo += 1                   # skip duplicate pair
    return result
```

## Complexity

| | Time | Space |
|---|---|---|
| Converging pair | O(n) | O(1) |
| 3Sum | O(n²) — sort + n passes | O(1) extra |

## Close-out ritual

Before you submit, say out loud:
1. Time and space complexity.
2. One edge case trace: two elements, all duplicates, or no valid answer.

## Plan problems

**A-set:** LC 125 · 167 · 15 · 11 · 42
**B-set:** LC 392 · 977 · 189 · 80