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

---

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        numset = sorted(nums)
        print(numset)
        ans = []
        for i in range(len(numset)-2):
            lp = i+1
            rp = len(numset)-1
            target = 0 - numset[i]
            temp = []
            while lp<rp:
                if numset[lp]+numset[rp] == target:
                    if [numset[i],numset[lp],numset[rp]] not in ans:
                        ans.append([numset[i],numset[lp],numset[rp]])
                    lp+=1
                    rp-=1
                elif numset[lp]+numset[rp] > target:
                    rp-=1
                elif numset[lp]+numset[rp] < target:
                    lp+=1
        return ans