3Sum

Problem Link

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 {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end()); // sort
        vector<vector<int>> result;
        
        for (int i = 0; i < nums.size() && nums[i] <= 0; ++i) {
            if (i > 0 && nums[i] == nums[i-1]) continue; // Skip duplicate values for nums[i]
            int left = i + 1, right = nums.size() - 1;

            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if (sum < 0) ++left;
                else if (sum > 0) --right;
                else {
                    result.push_back({nums[i], nums[left], nums[right]}); 
                    while (left < right && nums[left] == nums[left+1]) ++left; 
                    while (left < right && nums[right] == nums[right-1]) --right;

                    ++left; --right;
                }
            }
        }
        
        return result; // Return the list of unique triplets that sum up to 0
    }
};