0015-3sum

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Description

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

Solution(Python)

class Solution:
    def threeSum(self, nums: list[int]) -> list[list[int]]:
        # 
        # 
        # 
        #  Bruteforce n**3
        # for i 
        #     for j 
        #        for k 
        #          sum
        # a+ b +c = target
        # 
        # nums[k] = -(nums[i]+nums[j])
        # for every i
        #      run two sum
        #  complexity:(n**2)
        # 
        # 
        nums.sort()
        res = []
        n = len(nums)

        for i in range(n-2):
            if i > 0 and nums[i] == nums[i-1]:
                continue
            left = i + 1
            right = n - 1

            while left < right:
                total = nums[i] + nums[left] + nums[right]

                if total < 0:
                    left += 1
                elif total > 0:
                    right -= 1
                else:
                    res.append([nums[i], nums[left], nums[right]])
                    
                    left += 1
                    right -= 1

                    while left < right and nums[left] == nums[left - 1]:
                        left += 1

                    
                    # Skip duplicate right values
                    while left < right and nums[right] == nums[right + 1]:
                        right -= 1
                    

        return res

Dry Run