<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>

#include <iostream>
#include <vector>
#include <unordered_set>
#include <algorithm>

using namespace std;

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> result;
        int len = nums.size();
        if (len < 4) return result;
        sort(nums.begin(), nums.end());

        for (int i = 0; i < len - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;

            for (int j = i + 1; j < len - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) continue;

                unordered_set<long long> seen;
                for (int k = j + 1; k < len; k++) {
                    long long complement = (long long)target - (long long)nums[i] - (long long)nums[j] - (long long)nums[k];
                    if (seen.count(complement)) {
                        result.push_back({nums[i], nums[j], (int)complement, nums[k]});
                        while (k + 1 < len && nums[k] == nums[k + 1]) k++;
                    }
                    seen.insert(nums[k]);
                }
            }
        }

        return result;
    }
};
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