3sum

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = []
        nums.sort()
        for i in range(len(nums)):
            left = i + 1
            right = len(nums) - 1
            if nums[i] > 0:
                return res

            if i > 0 and nums[i] == nums[i - 1]:
                continue

            while left < right:
                if nums[i] + nums[right] + nums[left] > 0:
                    right -= 1
                elif nums[i] + nums[right] + nums[left] < 0:
                    left += 1
                elif nums[i] + nums[right] + nums[left] == 0:
                    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

3Sum

Difficulty: Medium

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