Question

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
• 105 <= nums[i] <= 105

Hint

If a + b + c = 0, then b + c = -a and c = -a - b

If we fix the value of a then we can loop through the remaining array for ‘b’ values and as a result ‘c’ (complement) values. Then we can increment the a value to the next element and repeat the loop.

Best Conceivable Runtime in this case is going to be O(n2)

Solution

My Solution:

def threeSum(self, nums: List[int]) -> List[List[int]]:
        seen, res = set(), set()
        d = {}
        # res = []
        for i,target in enumerate(nums):
            if target not in seen:
                seen.add(target)
                for j in nums[i+1:]:
                    complement = -target - j
                    if complement in d and d[complement] == i:
                        res.add(tuple(sorted([target, j, complement])))
                    d[j] = i
        return res