15. 3sum

Array Two Pointers Sorting

Problem - 3sum

Medium

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

Solutions

Python3

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        result = []

        for i in range(len(nums)):
            # skip dup nums
            if i > 0 and nums[i] == nums[i - 1]:
                continue

            target = -nums[i]
            left = i + 1
            right = len(nums) - 1
            while right > left:
                two_sum = nums[left] + nums[right]
                if two_sum == target:
                    result.append([nums[i], nums[left], nums[right]])
                    left += 1
                    right -= 1

                    #skip dups for left and right
                    while right > left and nums[left] == nums[left - 1]:
                        left += 1
                    while right > left and nums[right] == nums[right + 1]:
                        right -= 1

                elif two_sum < target:
                    left += 1
                else:
                    right -= 1
        return result

Submission Stats:

• Runtime: 483 ms (80.79%)
• Memory: 20.7 MB (70.58%)