<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:
    bool bin_search(vector<int>& sorted_arr, int start, int end, int target){
        if (start > end) return false; 
        int mid = (start + end) / 2;
        if (sorted_arr[mid] == target){
            return true;
        }
        if (sorted_arr[mid] < target){
            return bin_search(sorted_arr, mid + 1, end, target);
        }else{
            return bin_search(sorted_arr, start, mid - 1, target);
        }
    }
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        set<tuple<int,int,int>> s;
        for (int i = 0; i < nums.size() - 2; i ++){
            for (int j = i + 1; j < nums.size() - 1; j ++){
                int curr_sum = nums[i] + nums[j];
                if (bin_search(nums, j + 1, nums.size() - 1, -curr_sum)){
                    s.emplace(make_tuple(nums[i], nums[j], -curr_sum));
                }
            }
        }
        vector<vector<int>> ans;
        for (auto t : s){
            ans.push_back({get<0>(t), get<1>(t), get<2>(t)});
        }
        return ans;
    }
};
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