<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>& arr, int target) {
        sort(arr.begin(),arr.end());
        int n=arr.size();
        vector<vector<int>>ans;
        long long prev=INT_MAX;
        for(int i=0;i<n-3;i++){
            if(arr[i]==prev){
                continue;
            }
            long long sum=1LL*target-arr[i];
           
            prev=arr[i];
            long long prev2=INT_MAX;
            for(int j=i+1;j<n-2;j++){
                if(arr[j]==prev2){continue;}
                prev2=arr[j];
               long long temp=sum-arr[j];
               
                unordered_map<long,long>mp;
                int prev3=INT_MAX;
                 unordered_set<long long>st;
                for(int k=j+1;k<n;k++){
                 
                    if(mp.find(temp-arr[k])!=mp.end()){
                        
                        if(st.find(temp-arr[k])!=st.end()){
                            continue;
                        }
                        ans.push_back({arr[i],arr[j],(int)temp-arr[k],arr[k]});
                        st.insert(temp-arr[k]);
                    }
                    mp[arr[k]]++;
                    
                }
            }
        }
        return ans;
    }
};
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