3Sum - Leetcode 15 - Python

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.

class Solution:

    def threeSum(self, nums: List[int]) -> List[List[int]]:

        ans = []

        nums.sort()

        

        for i,a in enumerate(nums):

            if i>0 and a == nums[i-1]:

                continue

                

            l, r = i+1, len(nums)-1

            while l<r:

                tsum = a + nums[l] + nums[r]

                if tsum > 0:

                    r -= 1

                elif tsum < 0 :

                    l +=1 

                else:

                    ans.append([a,nums[l],nums[r]])

                    l += 1

                    while nums[l] == nums [l-1] and l < r:

                        l += 1

        return ans 

Explaination  (Hindi) :