3 Sum [Two Pointers]

Time: O(N^2); Space: O(1); medium

Given an array nums of N integers, are there elements a, b, c in nums such that a + b + c = 0?

Find all unique triplets in the array which gives the sum of zero.

Notes:

  • Elements in a triplet (a,b,c) must be in non-descending order. (ie, a <= b <= c)

  • The solution set must not contain duplicate triplets.

Example 1:

Input: nums = [-1, 0, 1, 2, -1, -4]

Output: [[-1, 0, 1], [-1, -1, 2]]

Hints:

  1. So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand!

  2. For the two-sum problem, if we fix one of the numbers, say ‘x’, we have to scan the entire array to find the next number ‘y’ which is value - x, where value is the input parameter. Can we change our array somehow so that this search becomes faster?

  3. The second train of thought for two-sum is, without changing the array, can we use additional space somehow? Like maybe a hash map to speed up the search?

[17]:

class Solution1(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums, result, i = sorted(nums), [], 0
        while i < len(nums) - 2:
            if i == 0 or nums[i] != nums[i - 1]:
                j, k = i + 1, len(nums) - 1
                while j < k:
                    if nums[i] + nums[j] + nums[k] < 0:
                        j += 1
                    elif nums[i] + nums[j] + nums[k] > 0:
                        k -= 1
                    else:
                        result.append([nums[i], nums[j], nums[k]])
                        j, k = j + 1, k - 1
                        while j < k and nums[j] == nums[j - 1]:
                            j += 1
                        while j < k and nums[k] == nums[k + 1]:
                            k -= 1
            i += 1
        return result

[18]:

s = Solution1()
nums = [-1, 0, 1, 2, -1, -4]
assert s.threeSum(nums) == [[-1, -1, 2], [-1, 0, 1]] or [[-1, 0, 1], [-1, -1, 2]]

[19]:

import collections

class Solution2(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        d = collections.Counter(nums)
        nums_2 = [x[0] for x in d.items() if x[1] > 1]
        nums_new = sorted([x[0] for x in d.items()])
        rtn = [[0, 0, 0]] if d[0] >= 3 else []
        for i, j in enumerate(nums_new):
            if j <= 0:
                numss2 = nums_new[i + 1:]
                for x, y in enumerate(numss2):
                    if 0 - j - y in [j, y] and 0 - j - y in nums_2:
                        if sorted([j, y, 0 - j - y]) not in rtn:
                            rtn.append(sorted([j, y, 0 - j - y]))
                    if 0 - j - y not in [j, y] and 0 - j - y in nums_new:
                        if sorted([j, y, 0 - j - y]) not in rtn:
                            rtn.append(sorted([j, y, 0 - j - y]))
        return rtn

[20]:

s = Solution2()
nums = [-1, 0, 1, 2, -1, -4]
assert s.threeSum(nums) == [[-1, -1, 2], [-1, 0, 1]] or [[-1, 0, 1], [-1, -1, 2]]