<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<pair<int, int>> twoSum(vector<int>& v, int start, int end, int target) {
        vector<pair<int, int>>res;
        unordered_map<string, bool> mp;
        int left = start;
        int right = end;
        while (left < right) {
            int currentTarget = v[left] + v[right];
            if (currentTarget == target) {
                pair<int, int>p;
                p.first=left;
                p.second=right;
                string s = to_string(left)+','+to_string(right);
                if(!mp[s]){
                    res.push_back(p);
                    mp[s]=true;
                }
                left++;
                right--;
            }
            else if (currentTarget > target)
                right--;
            else
                left++;
        }
        return res;
    }
    vector<vector<int>> threeSum(vector<int>& nums) {
        unordered_map<string, bool> mp;
        unordered_map<int, bool> sp;
        vector<vector<int>> res;
        sort(nums.begin(), nums.end());
        int n = nums.size();
        for (int i = 0; i < n - 2; i++) {
            if(sp[nums[i]]) continue;
            vector<pair<int,int>> ts = twoSum(nums, i + 1, n - 1, -1 * nums[i]);
            int m = ts.size();
            for(int j=0; j<m; j++){
                pair<int, int>pr = ts[j];
                string s = to_string(nums[i]) + ',' + to_string(nums[pr.first]) +
                       ',' + to_string(nums[pr.second]);
                if (!mp[s]) {
                    vector<int> v = {nums[i], nums[pr.first], nums[pr.second]};
                    res.push_back(v);
                    mp[s] = true;
                }
            }
            sp[nums[i]] = true;
        }
        return res;
    }
};
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