<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:
#define ll long long
    vector<vector<int>> fourSum(vector<int>& arr, int target) {
        set<vector<int>>ans ; 
        map<ll,vector<int>>ind ; 
        map<int,int>mpp; 
        for(int i  = 0 ; i<arr.size();i++){
            mpp[arr[i]]++ ; 
        }
        vector<int>nums; 
        for(auto it :  mpp){
            for(int j = 0 ; j <min(it.second , 4);j++){
                nums.push_back(it.first) ; 
            }
        }

        // ll target  = t ; 
        int n = nums.size() ; 
        for(int i = 0 ; i <n;i++) ind[nums[i]].push_back(i) ; 
        map<ll, vector<pair<int,int>>>right_sum ; 
        for(int i  = n-1;i>=0;i--){
            ll b = nums[i] ; 
            ind[nums[i]].pop_back(); 
            for(auto it :  right_sum){
                ll c_d =  it.first;
                for(auto it :  ind[target - b  - c_d]){
                    for(auto pp : right_sum[c_d]){
                        vector<int>temp = {nums[it] , nums[i] , nums[pp.first] , nums[pp.second]}  ; 
                        sort(temp.begin() , temp.end()) ; 
                        ans.insert(temp) ; 
                    }
                }
                
            }
            for(int j  = i+1 ; j <n;j++){
                right_sum[nums[i] +  nums[j]].push_back({i,j}) ; 
            }
        }
        vector<vector<int>>res; 
        for(auto it :ans) {
            res.push_back(it) ; 
        } 
        return res ;
        
    }
};
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