<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>

#include <bits/stdc++.h>
#define f(i, a, b) for (int i = a; i < b; i++)

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        
        int i, j, k, n = nums.size(), diff;
        sort(nums.begin(), nums.end());
        vector <vector <int>> ans;
        vector <int> temp;
        map <pair<int, int>, int> mp;

        f(i, 0, n - 2)
        {
            if (i != 0 && nums[i] == nums[i - 1]) continue;
            diff = 0 - nums[i];
            j = i + 1, k = n - 1;
            mp.clear();
            while (j < k)
            {
                if (mp[make_pair(nums[j], nums[k])] > 0) j++, k--;
                else {
                    temp.clear();
                    if (nums[j] + nums[k] == diff)
                    {
                        temp.push_back(nums[i]), temp.push_back(nums[j]), temp.push_back(nums[k]);
                        ans.push_back(temp);
                        mp[make_pair(nums[j], nums[k])]++;
                        j++, k--;
                    }
                    else if (nums[j] + nums[k] < diff) j++;
                    else k--;
                }
            }
        }
        return ans;
    }
};

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