<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 {
    typedef vector<int> triple;

public:
    vector<vector<int>> threeSum(vector<int>& nums_in) {
        auto nums = nums_in;
        set<vector<int>> solutionsSet;
        vector<set<int>> sub_solutions;
        auto const nums_count = nums.size();
        vector<vector<int>> solutions;
        sort(nums.begin(), nums.end());

        const int min_nums_i = nums[0];
        for (size_t j = 0; j < nums_count - 1; ++j) sub_solutions.push_back(std::set<int>());
        for (size_t j = 1; j < nums_count - 1; ) {
            const int max_nums_i = nums[j-1];
            // cout << "j=" << j << ", range: [" << min_nums_i << ", " << max_nums_i <<"]" << endl;
            std::set<int> sub_sol_j;
            size_t k = j + 1, k0 = k;
            size_t k1 = nums_count - 1;
            const int lower_bound_k =  -(nums[j] + max_nums_i);
            do {            
                //cout << k0 << " " << k << " " << k1 << " -> ";
                if (nums[k] > lower_bound_k) {
                    k1 = k;
                } else {
                    k0 = k;
                }
                k = (k0+k1) / 2;
                //cout << k0 << " " << k << " " << k1 << endl;
            } while (k < nums_count && k0 + 1 < k1);
            for (; k < nums_count;) {
                if (nums[j] + nums[k] + min_nums_i >0) break;
                if (!(nums[j] + nums[k] + max_nums_i <0)) {
                    // cout << "j=" << j << ", k=" << k << endl;
                    sub_sol_j.insert(nums[j] + nums[k]);
                }
                do {
                    ++k;
                } while (false && (k < nums_count) && (nums[k - 1] == nums[k]));
            }
            sub_solutions[j] = sub_sol_j;
            /*cout << "sub_solutions[" << j << "] <- ";
            for (auto s:sub_solutions[j]) {
                cout << s << ", ";
            }
            cout << endl;*/

            do {
                ++j;
            } while (false && (j < nums_count-1) && (nums[j - 1] == nums[j]));
        }

        for (size_t i = 0; i < nums_count - 2;) {
            for (size_t j = i + 1; j < nums_count - 1; ) {
                if (sub_solutions[j].count(-nums[i]) > 0) {
                    triple s;
                    s.push_back(nums[i]);
                    s.push_back(nums[j]);
                    s.push_back(-nums[i] - nums[j]);
                    solutionsSet.insert(s);
                    // solutionsSet.insert(*new triple(s));
                }
                do {
                    ++j;
                } while ((j < nums_count - 1) && (nums[j - 1] == nums[j]));
            }
            do {
                ++i;
            } while ((i < nums_count - 2) && (nums[i - 1] == nums[i]));
        }
        for (auto s : solutionsSet) {
           solutions.push_back(s);
        }
        /*
        for (auto s2 : sub_solutions) {
            for (auto sum2:s2.second) {
            triple s;
            s.push_back(s2.first);
            s.push_back(sum2);
            s.push_back(0);
            solutions.push_back(s);
            }
        }
        */

        return solutions;
    }
};
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