<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());
        vector<int>temp(3);
         vector<vector<int>>ans;
         int pos=-1;
         for(int i=0;i<nums.size();i++){
            if(nums[i]>=0){
                pos=i;
                break;
            }
         }
         if(pos>=0){
            for(int i=0;i<=pos;i++){
            for(int j=i+1;j<nums.size()-1;j++){
                int lo=pos;
                if(j>=pos) lo=j+1;
               int hi=nums.size()-1;
               while(lo<=hi){
                int mid=lo+(hi-lo)/2;
                if(nums[mid]==-(nums[i]+nums[j])){
                    temp[0]=nums[i];
                    temp[1]=nums[j];
                    temp[2]=nums[mid];
                    ans.push_back(temp);
                    break;
                }
                else if(nums[mid]>(-(nums[i]+nums[j]))) hi=mid-1;
                else lo=mid+1;
               }
            }
        }
        if(ans.size()>0){
             sort(ans.begin(),ans.end());
            stack<vector<int>>st;
            st.push(ans[0]);
        for(int i=1;i<ans.size();i++){
            if(st.top()!=ans[i]) st.push(ans[i]);
        }
        vector<vector<int>>finalAns;
        while(st.size()>0){
            finalAns.push_back(st.top());
            st.pop();
        }
        return finalAns;
        }
        else return ans;
         }
         else return ans;
        
    }
};
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