<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>& arr) {
        sort(arr.begin(),arr.end());
        int i=0;
        vector<vector<int>>sol;
       while(i<arr.size()-2){
        int j=i+1;int k=arr.size()-1;
        while(j<k){
             vector<int>ans(3);
            if(arr[i]+arr[j]+arr[k]==0){
                 ans[0]=arr[i];
                 ans[1]=arr[j];
                 ans[2]=arr[k];
                 sol.push_back(ans);
                 while(j<k && arr[j]==arr[j+1] ){
                    j++;
                 }
                 j++;
                 while(j<k && arr[k]==arr[k-1] ){
                    k--;
                 }
                 k--;
            }
            else if(arr[i]+arr[j]+arr[k]<0){
                while(j<k && arr[j]==arr[j+1] ){
                    j++;
                }
                j++;
            }
            else{
                 while(j<k && arr[k]==arr[k-1] ){
                    k--;
                }
                k--;
            }

        }
        while(i<arr.size()-1 && arr[i]==arr[i+1] ){
            i++;
        }
        i++;
        
       }
        return sol;

    }
};
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