<p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>

<ul>
	<li><code>0 &lt;= a, b, c, d&nbsp;&lt; n</code></li>
	<li><code>a</code>, <code>b</code>, <code>c</code>, and <code>d</code> are <strong>distinct</strong>.</li>
	<li><code>nums[a] + nums[b] + nums[c] + nums[d] == target</code></li>
</ul>

<p>You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,0,-1,0,-2,2], target = 0
<strong>Output:</strong> [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2,2,2,2], target = 8
<strong>Output:</strong> [[2,2,2,2]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 200</code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= target &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        unordered_map<int, int> mp;
        for(auto it: nums) mp[it]++;
        int n = nums.size();
        long long search;
        set<vector<int>> s;
        vector<vector<int>> ans;
        for (int i=0;i<n;i++) {
            for (int j=i+1;j<n;j++) {
                for (int k=j+1;k<n;k++){
                    mp[nums[i]]--, mp[nums[j]]--, mp[nums[k]]--;
                    search = long(target) - long(nums[i]) - long(nums[j]) - long(nums[k]);
                    if (search < INT_MAX && search > INT_MIN && mp.find(search) != mp.end() && mp[search] > 0) {
                        vector<int> temp = {nums[i], nums[j], nums[k], int(search)};
                        sort(temp.begin(), temp.end());
                        s.insert(temp);
                    }
                    mp[nums[i]]++, mp[nums[j]]++, mp[nums[k]]++;
                    // if (binary_search(nums.begin()+k, nums.end(), search))
                    //     ans.push_back({nums[i], nums[j], nums[k], search});
                }
            }
        }
        for (auto it:s) ans.push_back(it);
        return ans;
    }
};
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