3 Sum Leetcode Solution

#include 
#include 

using namespace std;

vector> threeSum(vector& nums) {
    vector> triplets;
    int n = nums.size();
    
    if (n < 3) {
        return triplets; // Not enough elements to form a triplet
    }
    
    sort(nums.begin(), nums.end()); // Sort the array in non-decreasing order
    
    for (int i = 0; i < n - 2; ++i) {
        // Skip duplicates for the first element
        if (i > 0 && nums[i] == nums[i - 1]) {
            continue;
        }
        
        int left = i + 1;
        int right = n - 1;
        
        while (left < right) {
            int total = nums[i] + nums[left] + nums[right];
            
            if (total == 0) {
                triplets.push_back({nums[i], nums[left], nums[right]});
                
                // Skip duplicates for the second element
                while (left < right && nums[left] == nums[left + 1]) {
                    left++;
                }
                // Skip duplicates for the third element
                while (left < right && nums[right] == nums[right - 1]) {
                    right--;
                }
                
                left++;
                right--;
            } else if (total < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
    
    return triplets;
}

class Solution {
    public List> threeSum(int[] nums) {
List<List<Integer>> triplets = new ArrayList<>();
        int n = nums.length;
        
        if (n < 3) {
            return triplets; // Not enough elements to form a triplet
        }
        
        Arrays.sort(nums); // Sort the array in non-decreasing order
        
        for (int i = 0; i < n - 2; ++i) {
            // Skip duplicates for the first element
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            int left = i + 1;
            int right = n - 1;
            
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total == 0) {
                    triplets.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip duplicates for the second element
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    // Skip duplicates for the third element
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    
                    left++;
                    right--;
                } else if (total < 0) {
                    left++;
                } else {
                    right--;
                }
            }
        }
        
        return triplets;    
    }
}

class Solution(object):
    def threeSum(self, nums):
    triplets = []
    nums.sort()  # Sort the array in non-decreasing order
    n = len(nums)
    
    for i in range(n - 2):
        # Skip duplicates for the first element
        if i > 0 and nums[i] == nums[i - 1]:
            continue
        
        left, right = i + 1, n - 1
        
        while left < right:
            total = nums[i] + nums[left] + nums[right]
            
            if total == 0:
                triplets.append([nums[i], nums[left], nums[right]])
                
                # Skip duplicates for the second element
                while left < right and nums[left] == nums[left + 1]:
                    left += 1
                # Skip duplicates for the third element
                while left < right and nums[right] == nums[right - 1]:
                    right -= 1
                
                left += 1
                right -= 1
            elif total < 0:
                left += 1
            else:
                right -= 1
    
    return triplets