# 4Sum C++ solution with explanation and comparison with 3Sum problem. Easy to understand.

• For the reference, please have a look at my explanation of `3Sum` problem because the algorithm are exactly the same. The link is as blow.

The key idea is to downgrade the problem to a `2Sum` problem eventually. And the same algorithm can be expand to `NSum` problem.

After you had a look at my explanation of `3Sum`, the code below will be extremely easy to understand.

``````class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {

vector<vector<int> > res;

if (num.empty())
return res;

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

for (int i = 0; i < num.size(); i++) {

int target_3 = target - num[i];

for (int j = i + 1; j < num.size(); j++) {

int target_2 = target_3 - num[j];

int front = j + 1;
int back = num.size() - 1;

while(front < back) {

int two_sum = num[front] + num[back];

if (two_sum < target_2) front++;

else if (two_sum > target_2) back--;

else {

// Processing the duplicates of number 3
while (front < back && num[front] == quadruplet[2]) ++front;

// Processing the duplicates of number 4
while (front < back && num[back] == quadruplet[3]) --back;

}
}

// Processing the duplicates of number 2
while(j + 1 < num.size() && num[j + 1] == num[j]) ++j;
}

// Processing the duplicates of number 1
while (i + 1 < num.size() && num[i + 1] == num[i]) ++i;

}

return res;

}
};
``````

• in the Nsum problem, your method will have N loops ........

• That's true. Nevertheless, I do remember that at somewhere I've read about the rigorous proof about this problem of which the result is bounded by `O(m^n)`, where `M` is the size of the array and `n` is the same number as the `n` in `n-th sum problem`. Correct me if I'm wrong.

• Is this O(nˆ3) runtime solution? it seems to be.

• yes it is cubic

• No actually it is bounded by O(m^n) where m is same as you stated but n is floor(n/2) in nth sum problem.

• @namangoyal, Could you show me the page where you see the conclusion? Thanks.

• num.empty() should be nums.size() < 4

• copy your solution with some changes

``````class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target)
{
vector<vector<int> > res;
if (num.size() < 4)
return res;
std::sort(num.begin(),num.end());
int i, j;
while (i < num.size() - 3)
{
j = i+1;
while (j < num.size() - 2)
{
int target_2 = target - num[i] - num[j];
int front = j + 1;
int back = num.size() - 1;
while(front < back) {
int two_sum = num[front] + num[back];
if (two_sum < target_2) front++;
else if (two_sum > target_2) back--;
else {
// Processing the duplicates of number 3
while (front < back && num[front] == quadruplet[2]) ++front;
// Processing the duplicates of number 4
while (front < back && num[back] == quadruplet[3]) --back;
} } // Processing the duplicates of number 2
while(j < num.size() - 2 && num[j + 1] == num[j++]);
} // Processing the duplicates of number 1
while (i < num.size() - 3 && num[i + 1] == num[i++]);
}
return res;
}
};``````

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