<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)
    {

        set<vector<int>> s;
        sort(nums.begin(), nums.end());
        vector<int> v;

        vector<vector<int>> vv;
        int size = nums.size();

        if (size < 3)
            return vv;

        // can't write here like if(nums[0]>0 && size<3) return vv;
        // cause if size is zero(empty vector) then here you are checking on nums[0] >0... 
        // out of bound memory runtime error


        if (nums[0] > 0)
            return vv; 
        //if first element in sorted array is greater than 0 
        //then no way triplet can be 0
        

        int j, k;
        for (int i = 0; i < size - 2; i++)
        {
            j = i + 1, k = size - 1;
            while (j < k)
            {
                if (nums[j] + nums[k] == -nums[i])
                {
                    v.push_back(nums[i]);
                    v.push_back(nums[j]);
                    v.push_back(nums[k]);
                    s.insert(v);
                    v.clear();
                }
                if (nums[j] + nums[k] > -nums[i])
                    k--;
                else
                    j++;
            }
        }
        for (auto i = s.begin(); i != s.end(); i++)
        {
            vv.push_back(*i);
        }
        return vv;
    }
};
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