Saved from https://leetcode.com/problems/3sum/submissions/

lc15

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        """
        Three pointer solution built on top of the two-sum problem. 
        Time: O(n^2)
        Space: O(1)
        The trick is how to handle duplicate. The order of checking two pointers' equality matters: one must move pointer first, 
        then check equality against the prior value. The case starting from the prior value included the second case.  
        E.g. -1, 0, 0, 1
        0, 0, 1 included the case of 0, 1. But the case starts from the second 0 doesn't include the prior case.
        """
        nums.sort()
        res = []
        n = len(nums)
        for start in range(n-1):
            l, r = start+1, n-1
            if start > 0 and nums[start] == nums[start-1]:
                continue    # remove duplicates
            while l < r:
                if nums[start] + nums[l] + nums[r] == 0:
                    res.append([nums[start], nums[l], nums[r]])
                    while l < r:
                        l += 1
                        if nums[l] != nums[l-1]:
                            break
                    while l < r:
                        r -= 1
                        if nums[r] != nums[r+1]:
                            break
                elif nums[start] + nums[l] + nums[r] > 0:
                    r -= 1
                elif nums[start] + nums[l] + nums[r] < 0:
                    l += 1
        return res