# 3Sum

### [3Sum]

Given an array S of n integers, are there elements abc in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

• Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
• The solution set must not contain duplicate triplets.

For example, given array S = {-1 0 1 2 -1 -4}, A solution set is: (-1, 0, 1) (-1, -1, 2)

### [Analysis]

The most naive approach would be $O(n^2)$. Note that the sum is 0, if we sort the array at the beginning, we can save a lot of useless check.

There are several optimizations to make though, more details commented in the code.

vector<vector<int> > threeSum(vector<int> &num) {
vector<vector<int> > res;

if (num.size() < 3)
return res;

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

for (int i=0; i<num.size()-2; ++i) {
// there is no need to go further if num[i] doesn't change.
// Suppose that we've found a new valid triplet,
// it must be a duplicate because previous start and end are already included with this num[i]
// Also, this gurantee that if num[i] has increased,
// this won't be a duplicate, since we know that num[start] and num[end] are both greater than num[i]
if (i>0 && num[i] == num[i-1]) continue;
// we know that num[i] <= num[start] <= num[end],
// if num[i] > 0 than there is no hope to find a valid triplet.
if (num[i] > 0) break;

int start = i+1, end = num.size()-1;

while (start < end) {
int s = num[start]+num[end]+num[i];
if(start>i+1 && num[start] == num[start-1]) {
start++;
continue; // Skip the same num[j] here.
}

if (s > 0) {
end--;
} else if (s < 0){
start++;
} else {
// something new since num[i] has increased
vector<int> r = {num[i], num[start], num[end]};
res.push_back(r);
start++;
end--;
}
}
}

return res;
}