<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>> twosum(vector<int>&nums, int cur) {
        int n = nums.size();
        int st = cur + 1;
        int en = n - 1;
        int target = -nums[cur];
        vector<vector<int>> resAll;
        while(st < en) {
            if(nums[st] + nums[en] < target) {
                st++;
            }
            else if(nums[st] + nums[en] > target) {
                en--;
            }
            else {
                vector<int> res;
                res.push_back(nums[st]);
                res.push_back(nums[en]);
                res.push_back(nums[cur]);
                resAll.push_back(res);
                // break;
                while(st < n - 1 && nums[st] == nums[st + 1]) st++;
                while(en > 0 && nums[en] == nums[en - 1]) en--; 
                st++;en--;

            }
        }
        return resAll;
    }

    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> resAll;
        int n = nums.size();
        for(int i = 0; i < n; i++) {
            if(i > 0 && nums[i] == nums[i - 1]) continue;
            vector<vector<int>> res = twosum(nums, i);
            if(!res.empty()) {
                // res.push_back(nums[i]);
                int in = resAll.size();
                
                // if(!resAll.empty()){
                //     in--;
                // cout << resAll[in][0];
                // cout << resAll[in][1];
                // cout << resAll[in][2];
                //     if(resAll[in][0] != res[0] && resAll[in][1] != res[1])
                //         resAll.push_back(res);
                //     }
                // else 
                resAll.reserve(resAll.size() + res.size());
                resAll.insert( resAll.end(), res.begin(), res.end());
                // resAll.push_back(res);
            }
        }
        return resAll;

    }
};
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