<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) {
             sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<vector<int>> ans; int i=0;
    


        while(i<n){
            int a = nums[i];
            vector<int> temp;
            temp.push_back(a);
                    int prevB=INT_MAX; 
                  
           
           

            for(int j=i+1; j<n; j++){
               int b = nums[j];
               if(b==prevB)continue;
               prevB=b;
            map<long, int> mp;
          long restTarget=(long)target-(a+b);
            temp.push_back(b);
              int prevC=INT_MAX;

            for(int k =j+1; k<n; k++){
                if(nums[k]==prevC) continue;
                if(mp.find(restTarget-nums[k])!= mp.end()){
                    mp[restTarget-nums[k]]--;
                    if(mp[restTarget-nums[k]]==0) mp.erase(restTarget-nums[k]);
                    temp.push_back(nums[k]); prevC=nums[k];
                    temp.push_back(restTarget-nums[k]);
                    ans.push_back(temp);
                    temp.pop_back(); temp.pop_back();
                }
                mp[nums[k]]++;
            }
            temp.pop_back();
            }
             
            while(i<n && nums[i]==a){i++;}


            }
            return ans;

    }
};
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