15. 3Sum

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != ji != 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]]

Example 2:

Input: nums = []
Output: []

Example 3:

Input: nums = [0]
Output: []


  • 0 <= nums.length <= 3000
  • -105 <= nums[i] <= 105


Sort based algorithm

  • a+b = -c. 3SUM reduces to 2SUM problem.

Handling Duplicates in 2SUM

  • Say index s and e are forming a solution in a sorted array. Now givens nums[s], there is a unique nums[e] such that nums[s]+nums[e]=Target. Therefore, if nums[s+1] is the same as nums[s], then searching in range s+1 to e will give us a duplicate solution. Thus we must move s till nums[s] != nums[s-1] to avoid getting duplicates.
while s<e and nums[s] == nums[s-1]:
      s = s+1

Handling Duplicates in 3SUM

  • Imagine we are at index i and we have invoked the 2SUM problem from index i+1 to end of the array. Now once the 2SUM terminates, we will have a list of all triplets which include nums[i]. To avoid duplicates, we must skip all nums[i] where nums[i] == nums[i-1].
if i > 0 and nums[i] == nums[i-1]:


class Solution(object):
    def threeSum(self, nums):
        :type nums: List[int]
        :rtype: List[List[int]]
        N, result = len(nums), []
        for i in range(N):
            if i > 0 and nums[i] == nums[i-1]:
            target = nums[i]*-1
            s,e = i+1, N-1
            while s<e:
                if nums[s]+nums[e] == target:
                    result.append([nums[i], nums[s], nums[e]])
                    s = s+1
                    while s<e and nums[s] == nums[s-1]:
                        s = s+1
                elif nums[s] + nums[e] < target:
                    s = s+1
                    e = e-1
        return result

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