15. 3-sum

Constraints

1. 3 <= nums.length <= 3000
2. -105 <= nums[i] <= 105

Idea #1 (Time: O(N), Memory: O(N))

1. sort nums
2. iterate if len(nums) > 2
3. check duplicate
4. left, right setup
5. while left < right
6. check if 3sum < 0: left++, >0: right--
7. 3sum == 0: append 3 elem @ res and check duplicate both left and right, left++, right--
8. if done return res, if not go to 2

Test Cases

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.

Code

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = list()
        nums.sort()
        for i in range(len(nums) - 2):
            if i > 0 and nums[i] == nums[i-1]:
                continue
            left, right = i + 1, len(nums) - 1
            while left < right:
                sum = nums[i] + nums[left] + nums[right]
                if sum < 0:
                    left += 1
                elif  sum > 0:
                    right -= 1
                else:
                    res.append([nums[i], nums[left], nums[right]])
                    while left < right and nums[left] == nums[left + 1]:
                        left += 1
                    while left < right and nums[right] == nums[right -1]:
                        right -= 1
                    left += 1
                    right -= 1
        return res