<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:
    // ios::sync_with_stdio(0);
    //   cin.tie(0);
    //   cout.tie(0);
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        unordered_map<long long, int> m;
        map<vector<int>, int> m1;
        vector<vector<int>> v;
        long long tg = target;
        for (int i = 0; i < nums.size(); i++) {
            m[nums[i]]++;
        }
        if (nums.size() < 4) {
            return v;
        }
        for (int i = 0; i < nums.size() - 3; i++) { m[nums[i]]--;
            for (int j = i + 1; j < nums.size() - 2; j++) {  m[nums[j]]--;
                for (int k = j + 1; k < nums.size()-1; k++) {  m[nums[k]]--;
                    long long tp = nums[k];
                    tp += (long long)nums[i];
                    tp += (long long)nums[j];
                    tp = (long long)tg - tp;
                    // cout<<tp<<endl;
                    if (m[tp]) {
                        vector<int> ans;
                        ans.push_back({nums[i]});
                        ans.push_back({nums[j]});
                        ans.push_back({nums[k]});
                        ans.push_back({(int)tp});
                        sort(ans.begin(), ans.end());
                        if (m1[ans] == 0) {
                            v.push_back(ans);
                            m1[ans]++;
                        }
                    }
                    m[nums[k]]++;
                }
                m[nums[j]]++;
            }
            m[nums[i]]++;
        }
        return v;
    }
};
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