Day028 - 15. 3Sum

15. 3Sum

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

내 풀이

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        answer = []
        
        for fix in range(len(nums) - 2):
            if fix > 0 and nums[fix] == nums[fix - 1]:  # 고정값 중복 방지
                continue

            l, r = fix + 1, len(nums) - 1
            while l < r:
                three_sum = nums[fix] + nums[l] + nums[r]

                if three_sum == 0:
                    answer.append([nums[fix], nums[l], nums[r]])
                    
                    # 중복된 값 건너뛰기
                    while l < r and nums[l] == nums[l + 1]:
                        l += 1
                    while l < r and nums[r] == nums[r - 1]:
                        r -= 1
                    
                    l += 1
                    r -= 1
                elif three_sum < 0:
                    l += 1
                else:
                    r -= 1

        return answer

# Time Complexity : \(O(N)\)