# 3Sum Problem & Solution

See the 3sum problem on LeetCode.

## C++ Solution

The runtime complexity of this algorithm is \$O(n\log{}n + n^2)\$.

``````#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
#pragma GCC target("avx,avx2,fma")

static const int _=[](){ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();

class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> result;

sort(nums.begin(), nums.end());

for (int i = 0; i < (int)nums.size() - 2; ++i) {
int j = i + 1;
int k = nums.size() - 1;
while (j < k) {
int sum = nums[i] + nums[j] + nums[k];
if (sum > 0) {
—k;
} else if (sum < 0) {
++j;
} else {
result.push_back({nums[i], nums[j], nums[k]});
while (j < (int)nums.size() - 1 && nums[j] == nums[j + 1]) ++j;
while (k > 0 && nums[k] == nums[k - 1]) —k;
++j;
—k;
}
}

while (i < (int)nums.size() - 1 && nums[i] == nums[i + 1]) ++i;
}

return vector<vector<int>>(result.begin(), result.end());
}
};
``````

## Start Here

Many paths, there are. Follow yours, you must.