<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) {
        vector<vector<int>> ans;
        sort(nums.begin(), nums.end());
        map<int, bool> done;
         map<vector<int>, bool> doneSub;
        for (int i = 0; i < nums.size(); i++) {  
                if(done.find(nums[i])==done.end()){
                done[nums[i]] = true;
                int lo = i+1;
                int hi = nums.size() - 1;
                int n = nums[i];
                map<int,bool> tempDone ={};
                while (lo < hi) {
                    if (lo == i) {
                        lo += 1;
                    } 
                    else if (hi == i) {
                        hi -= 1;
                    }
                     else {
                        if(tempDone.find(nums[lo])!=tempDone.end()){    
                            lo+=1;
                        }
                        else if (tempDone.find(nums[hi])!=tempDone.end()){
                            hi-=1;
                        }
                        else{
                        
                        int sum = nums[lo] + nums[hi] + n;
                        if (sum == 0) {
                            vector<int> cAns;
                            cAns.push_back(nums[lo]);
                            cAns.push_back(nums[hi]);
                            cAns.push_back(n);
                            // sort(cAns.begin(), cAns.end());
                            tempDone[nums[hi]] = true;
                            tempDone[nums[lo]] = true;
                            lo+=1;
                            hi-=1;
                            if (doneSub.find(cAns) == doneSub.end()) {
                                ans.push_back(cAns);
                                doneSub[cAns] = true;
                            }
                            
                        } else if (sum < 0) {
                            lo += 1;
                        } else {
                            hi -= 1;
                        }
                        }
                    }
                }
            
        }}
        return ans;
    }
};
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