3 Sum

15. 3 Sum

Difficulty: Medium

Related Topics: Array, Two Pointers, Sorting

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.

Code:

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = []
        nums.sort()

        for ind, num in enumerate(nums):
            if ind > 0 and nums[ind-1] == num:
                continue
            l, r = ind+1, len(nums)-1
            while l < r:
                three = nums[l] + nums[r] + num
                if three == 0:
                    res.append([nums[l], nums[r], num])
                    l += 1
                    while nums[l] == nums[l-1] and l < r:
                        l += 1
                elif three > 0:
                    r-=1
                else:
                    l+=1
        return res