<p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>

<p>Notice that the solution set must not contain duplicate triplets.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
<strong>Output:</strong> [[-1,-1,2],[-1,0,1]]
<strong>Explanation:</strong> 
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,1]
<strong>Output:</strong> []
<strong>Explanation:</strong> The only possible triplet does not sum up to 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,0,0]
<strong>Output:</strong> [[0,0,0]]
<strong>Explanation:</strong> The only possible triplet sums up to 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= nums.length &lt;= 3000</code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<vector<int>> ans;
        if(nums[0]==0 && nums[nums.size()-1]==0) {
            ans.push_back({0, 0, 0});
            return ans;
        }
        for(int i=0; i<nums.size(); i++) {
            int x = i+1, y=n-1;
            while(x<y && x<n && y>=0) {
                if(nums[i]+nums[x]+nums[y]==0) {
                    ans.push_back({nums[i], nums[x], nums[y]});    
                    x++;   
                } else if(nums[x]+nums[y] > -nums[i]) {
                    y--;
                } else {
                    x++;
                }
            }
            if(nums[i]>0) break;
        }
        sort(ans.begin(), ans.end());
        vector<vector<int>> ans1;
        if(ans.size()==0) return ans1;        
        ans1.push_back(ans[0]);
        int ptr=0;
        for(int i=1;i<ans.size();i++) {
            if(ans[i]!=ans1[ptr]) {
                ans1.push_back(ans[i]);
                ptr++;
            }
        }
        return ans1;
    }
};
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