<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) {
        int n = nums.size();
        vector<int> voidvector({});
        vector<vector<int>> ans(1, voidvector);
        ans.clear();
        if (n < 4) {
            return ans;
        }
        int x = 0;
        while (x < n){
            int coount = count(nums.begin(), nums.end(), nums[x]);
            if(coount> 4){
                int xd = coount - 5;
                for(int i = (n-1); i>0; i--){
                    if(nums[i]==nums[x]){
                        nums.erase(nums.begin()+i);
                        coount--;
                        break;
                    }
                }
            }
            n = nums.size();
            x++;
        }
        for (int i = 0; i < (n - 3); i++) {
            for (int j = (i + 1); j < (n - 2); j++) {
                for (int k = (j + 1); k < (n - 1); k++) {
                    for (int m = (k + 1); m < n; m++) {
                        long long sum = nums[i];
                        sum = sum + nums[j] + nums[k] + nums[m];
                        if (sum == target) {
                            vector<int> quadruplet(
                                {nums[i], nums[j], nums[k], nums[m]});
                                sort(quadruplet.begin(),quadruplet.end());
                            if (count(ans.begin(), ans.end(), quadruplet) ==
                                0) {
                                ans.push_back(quadruplet);
                            }
                        }
                    }
                }
            }
        }
        return ans;
    }
};
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