<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>

using ll = long long;
class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        vector<vector<int>> ans;
        for (int a = 0; a < n; a++) {
            if (ll(nums[a]) * 4LL > target) break;
            if (a != 0 && nums[a-1] == nums[a]) continue;
            for (int b = a+1; b < n; b++) {
                if (ll(nums[a]) + ll(nums[b]) * 3LL > target) break;
                if (b != a+1 && nums[b-1] == nums[b]) continue;
                unordered_set<ll> seen;
                for (int d = b+1; d < n; d++) {
                    ll soFar = ll(nums[a]) + ll(nums[b]) + ll(nums[d]);
                    ll want = ll(target) - soFar;
                    if (seen.count(want) == 1) {
                        ans.push_back({nums[a], nums[b], int(want), nums[d]});
                    }
                    seen.insert(ll(nums[d]));
                }
            }
        }
        if (ans.empty()) return {};
        sort(ans.begin(), ans.end());
        vector<vector<int>> ans2{ans[0]};
        for (int i = 1; i < ans.size(); i++) {
            if (ans[i-1] == ans[i]) continue;
            ans2.push_back(ans[i]);
        }
        return ans2;
    }
};
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