<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) {
        unordered_map<long long, set<pair<int, int>>> prevPairs;

        vector<vector<int>> result;
        set<vector<int>> addedQuads;

        for (int i = 0; i < nums.size(); i++) {
            long long icomp = target - nums[i];
            unordered_set<int> visited;

            for (int j = i + 1; j < nums.size(); j++) {
                long long complement = icomp - nums[j];
                auto& s = prevPairs[complement];
                if (visited.contains(nums[j]))
                    continue;
                auto it = prevPairs.find(complement);
                if (it != prevPairs.end()) {
                    for (auto [n1, n2] : it->second) {
                        vector<int> v { n1, n2, nums[i], nums[j] };
                        sort(v.begin(), v.end());
                        if (addedQuads.contains(v)) continue;
                        result.push_back(v);
                        addedQuads.insert(v);
                    }
                }
            }

            for (int j = 0; j < i; j++) {
                int sum = nums[i] + nums[j];
                prevPairs[sum].insert({ nums[i], nums[j] });
            }
        }

        return result;
    }
};
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