<p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>

<p>Notice that the solution set must not contain duplicate triplets.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
<strong>Output:</strong> [[-1,-1,2],[-1,0,1]]
<strong>Explanation:</strong> 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,1]
<strong>Output:</strong> []
<strong>Explanation:</strong> The only possible triplet does not sum up to 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,0,0]
<strong>Output:</strong> [[0,0,0]]
<strong>Explanation:</strong> The only possible triplet sums up to 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= nums.length &lt;= 3000</code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

// Jai shree ram
// Hash function for vector<int>
struct VectorHash {
    size_t operator()(const vector<int>& v) const {
        size_t hash_value = 0;
        for (int i : v) {
            hash_value ^= hash<int>()(i) + 0x9e3779b9 + (hash_value << 6) + (hash_value >> 2);
        }
        return hash_value;
    }
};

map<vector<int>, vector<vector<int>>> pre;

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& v) {
        int n = v.size();
        sort(v.begin(), v.end());
        if(pre.find(v) != pre.end())
        {
            return pre[v];
        }

        unordered_map<int, int> mp;
        for (int it : v) mp[it]++;

        set<vector<int>> s;
        

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int req = 0 - (v[i] + v[j]);
                if (req < 0) break;
                mp[v[i]]--;
                mp[v[j]]--;
                if (mp[req] > 0) {
                    vector<int> te = {v[i], v[j], req};
                    sort(te.begin(), te.end());
                    s.insert(te);
                }
                mp[v[i]]++;
                mp[v[j]]++;
            }
        }

        vector<vector<int>> ans(s.begin(), s.end()); 
        pre[v] = ans;
        return ans;

    }
};
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