<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) {
        vector<vector<int>>ans;
        unordered_map<long long,long long>m;
        long long n = nums.size();
        for(int i=0;i<n; i++){
            m[nums[i]] = i;
        }
        set<vector<int>>s;
        //sort(nums.begin(), nums.end());
        for(int i=0; i<n; i++){
            for(int j=i+1; j<n; j++){
                for(int k=j+1; k<n; k++){
                    long long f =  (long long)target-nums[i]-nums[j]-nums[k];
                    if(m.find(f)!=m.end() && m[f]!=i && m[f]!=j && m[f]!=k){
                        vector<int>v;
                        v.push_back(nums[i]);
                        v.push_back(nums[j]);
                        v.push_back(nums[k]);
                        v.push_back(nums[m[f]]);
                        sort(v.begin(), v.end());
                        auto it = s.find(v);
                        s.insert(v);
                        if(it == s.end())
                        ans.push_back(v);
                    }
                }
            }
        }
        return ans;
    }
};
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