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15. 3Sum

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

Array

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

Solution

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

Last updated on September 24, 2026

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