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MEDIUM

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

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].

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Constraints

• 3 ≤ nums.length ≤ 3000
• -10⁵ ≤ nums[i] ≤ 10⁵

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Solution: Two Pointers after Sorting

• Time Complexity: O(n²), where n is the length of the array.
• Space Complexity: O(|ans|), where |ans| is the number of triplets in the answer.

C++

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> ans;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size(); i++) {
            if(i > 0 && nums[i] == nums[i-1]) continue;
            int j = i + 1, k = nums.size()-1;
            while(j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                if (sum == 0) {
                    ans.push_back({nums[i], nums[j++], nums[k--]});
                    while(j < nums.size() && nums[j] == nums[j-1]) j++;
                    while(k >= 0 && nums[k] == nums[k+1]) k--;
                }
                else if (sum < 0) j++;
                else k--; // sum > 0
            }
        }
        return ans;
    }
};