ThreeSum

Problem

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.

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:sup:`5`<= nums[i] <= 10:sup:`5`

Solution

Sort nums. We can apply the 2Sum solution by iterating i from 0 to len(nums) - 1 through nums and finding any remaining 2 numbers nums[j], nums[k] such that all 3 sum to 0. Track the solutions in a set to avoid duplicates. Moreover, if nums[i] is a repeat, we can continue to the next iteration.

Pattern

Array, Two Pointers, Sorting

Code

from typing import List


def threeSum(nums: List[int]) -> List[List[int]]:
    """Returns a list of triplets in ``nums`` with distinct indices that sum to
    0.
    """
    solutions = set()
    nums = sorted(nums)
    for i, num in enumerate(nums):
        if i > 0 and nums[i] == nums[i - 1]:
            continue

        left = i + 1
        right = len(nums) - 1
        while left < right:
            s = nums[i] + nums[left] + nums[right]
            if s < 0:
                left += 1
            elif s > 0:
                right -= 1
            else:
                solutions.add((nums[i], nums[left], nums[right]))
                left += 1
                right -= 1
    return [list(solution) for solution in solutions]

Test

>>> from ThreeSum import threeSum
>>> threeSum([-1, 0, 1, 2, -1, -4])
[[-1, 0, 1], [-1, -1, 2]]
>>> threeSum([0, 1, 1])
[]
>>> threeSum([0, 0, 0])
[[0, 0, 0]]

ThreeSum.threeSum(nums: List[int]) → List[List[int]]: Returns a list of triplets in nums with distinct indices that sum to 0.