0015: 3Sum

Problem Statement

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.

Constraints:

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

Code Solution

class Solution:
    def threeSum(self, nums):
	    N = len(nums)
	    
        nums.sort()
        
        res = []
        for ix in range(N - 2):
            if ix > 0 and nums[ix - 1] == nums[ix]:
                continue
            jx = 1 + ix
            kx = N - 1
            while jx < kx:
                s = nums[ix] + nums[jx] + nums[kx]
                if s < 0:
                    jx = jx + 1
                elif s > 0:
                    kx = kx - 1
                else:
                    res.append([
                        nums[ix],
                        nums[jx],
                        nums[kx]
                    ])
                    while jx < kx and nums[kx] == nums[kx - 1]:
                        kx = kx - 1
                    kx = kx - 1
        return res