---
title: '15. 3Sum'
description: Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0
sidebar:
  label: '3Sum'
  badge: 'Medium'
---

Array

::::warning
Notice that the solution set must not contain duplicate triplets.
::::

### Example 1:
- Input: `nums = [-1,0,1,2,-1,-4]`
- Output: `[[-1,-1,2],[-1,0,1]]`
- Explanation: `nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0`. `nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0`. `nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0`. The distinct triplets are `[-1,0,1]` and `[-1,-1,2]`. Notice that the order of the output and the order of the triplets does not matter.

### Example 2:
- Input: `nums = [0,1,1]`
- Output: `[]`
- Explanation: The only possible triplet does not sum up to `0`.

### Example 3:
- Input: `nums = [0,0,0]`
- Output: `[[0,0,0]]`
- Explanation: The only possible triplet sums up to `0`.

### Constraints:

- `3 <= nums.length <= 3000`
- `-10^5 <= nums[i] <= 10^5`

## Approach

```mermaid
flowchart TD
  S(["threeSum(nums)"]) --> O["nums.sort() — duplicates become adjacent"]
  O --> F{"more i in 0..n-1?"}
  F -- no --> E(["return result"])
  F -- yes --> D{"nums[i] == nums[i-1]?"}
  D -- yes --> F
  D -- no --> P["left = i+1, right = n-1, target = -nums[i]"]
  P --> W{"left < right?"}
  W -- no --> F
  W -- yes --> C["current = nums[left] + nums[right]"]
  C --> Q{"current vs target"}
  Q -- equal --> A["append triplet, skip equal neighbours, left += 1, right -= 1"]
  A --> W
  Q -- "current < target" --> L["left += 1 — need a bigger sum"]
  L --> W
  Q -- "current > target" --> R["right -= 1 — need a smaller sum"]
  R --> W
```

## Solution

```py
class Solution:
    def threeSum(self, nums: list[int]) -> list[list[int]]:
        nums.sort()
        result = []
        n = len(nums)

        for i in range(n):
            # skip all zero
            if i > 0 and nums[i] == nums[i - 1]:
                continue

            # two pointers
            left = i + 1
            right = n - 1
            target = -nums[i]

            while left < right:
                current = nums[left] + nums[right]

                if current == target:
                    result.append([nums[i], nums[left], nums[right]])

                    # skip duplicates
                    while left < right and nums[left] == nums[left + 1]:
                        left += 1
                    while left < right and nums[right] == nums[right - 1]:
                        right -= 1

                    # shift pointers
                    left += 1
                    right -= 1

                # since sorted, if current < target, then move left
                elif current < target:
                    left += 1
                # since sorted, if current > target, then move right
                else:
                    right -= 1

        return result
```

## Explanation

[3Sum - Leetcode 15 - Python](https://www.youtube.com/watch?v=jzZsG8n2R9A)