the key idea is the same as the `TwoSum`

problem. When we fix the `1st`

number, the `2nd`

and `3rd`

number can be found following the same reasoning as `TwoSum`

.

The only difference is that, the `TwoSum`

problem of LEETCODE has a unique solution. However, in `ThreeSum`

, we have multiple duplicate solutions that can be found. Most of the OLE errors happened here because you could've ended up with a solution with so many duplicates.

The naive solution for the duplicates will be using the STL methods like below :

```
std::sort(res.begin(), res.end());
res.erase(unique(res.begin(), res.end()), res.end());
```

But according to my submissions, this way will cause you double your time consuming almostly.

A better approach is that, to jump over the number which has been scanned, no matter it is part of some solution or not.

If the three numbers formed a solution, we can safely ignore all the duplicates of them.

We can do this to all the three numbers such that we can remove the duplicates.

Here's my AC C++ Code:

```
vector<vector<int> > threeSum(vector<int> &num) {
vector<vector<int> > res;
std::sort(num.begin(), num.end());
for (int i = 0; i < num.size(); i++) {
int target = -num[i];
int front = i + 1;
int back = num.size() - 1;
while (front < back) {
int sum = num[front] + num[back];
// Finding answer which start from number num[i]
if (sum < target)
front++;
else if (sum > target)
back--;
else {
vector<int> triplet(3, 0);
triplet[0] = num[i];
triplet[1] = num[front];
triplet[2] = num[back];
res.push_back(triplet);
// Processing duplicates of Number 2
// Rolling the front pointer to the next different number forwards
while (front < back && num[front] == triplet[1]) front++;
// Processing duplicates of Number 3
// Rolling the back pointer to the next different number backwards
while (front < back && num[back] == triplet[2]) rear--;
}
}
// Processing duplicates of Number 1
while (i + 1 < num.size() && num[i + 1] == num[i])
i++;
}
return res;
}
```