Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

a + b + c = 0

Solution is to just convert this into two pointer technique of solving 2SUM problem. Traverse the array and for each element check if rest of the array has 2 elements that sum up to -a.

class Solution(object):     def threeSum(self, nums):         def two_pointer(nums, target):             '''two pointer technique and it is caller responsibility to pass proper nums'''             first = 0             second = len(nums) - 1             two_sums = []             while first < second:                 two_sum = nums[first] + nums[second]                 if two_sum < target:                     first += 1                 elif two_sum > target:                     second -= 1                 else:                     two_sums.append([nums[first]] + [nums[second]])                     while first+1 < len(nums) and nums[first] == nums[first+1]:                         first += 1                     while second-1 >= 0 and nums[second] == nums[second-1]:                         second -= 1                     first += 1                     second -= 1             return two_sums         """         :type nums: List[int]         :rtype: List[List[int]]         """         nums = sorted(nums)         three_sums = []         i = 0         while i < len(nums)-2:             two_sums = two_pointer(nums[i+1:], -nums[i])             for j in range(len(two_sums)):                 three_sums.append([nums[i]] + two_sums[j])             while i+1 < len(nums) and nums[i] == nums[i+1]:                 i += 1             i += 1         return three_sums