LeetCode - 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

 

ChatGPT's Solution:

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        
        nums.sort() # Sort the array to make it easier to find triplets and avoid duplicates
        n = len(nums)
        res = []

        for i in range(n - 2):
            # Skip duplicate numbers for the first element
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            
            left, right = i + 1, n - 1  # Two-pointer approach

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

                if total == 0:
                    res.append([nums[i], nums[left], nums[right]])

                    # Skip duplicate numbers for the second and third elements
                    while left < right and nums[left] == nums[left + 1]:
                        left += 1
                    while left < right and nums[right] == nums[right - 1]:
                        right -= 1
                    
                    # Move pointers inward
                    left += 1
                    right -= 1
                elif total < 0:
                    left += 1  # Increase sum by moving left pointer
                else:
                    right -= 1  # Decrease sum by moving right pointer

        return res

 

Explanation:

1. Sorting: Sorting the array allows us to efficiently find triplets while avoiding duplicates.
2. Iterate over each element (nums[i]) as a fixed starting number:
   • Skip duplicates to avoid repeated triplets.
3. Use two pointers (left and right):
   • Move left and right to find a sum of 0.
   • Skip duplicate values to prevent duplicate triplets in the result.
4. Time Complexity: O(n²) due to sorting (O(n log n)) and two-pointer traversal (O(n) per fixed i).