216 Combination Sum III

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Example 1:
Input: k = 3, n = 7
Output:
[[1,2,4]]
Example 2:
Input: k = 3, n = 9
Output:
[[1,2,6], [1,3,5], [2,3,4]]
• Very icky, brute force approach of finding all the valid combinations and testing them.
vector<int> people;
vector<int> combination;
vector<vector<int>> final_combs;
void pretty_print(const vector<int>& v) {
final_combs.push_back(v);
//static int count = 0;
//cout << "combination no " << (++count) << ": [ ";
//for (int i = 0; i < v.size(); ++i) { cout << v[i] << " "; }
//cout << "] " << endl;
//pause();
}
void go(int offset, int k) {
if (k == 0) {
pretty_print(combination);
return;
}
for (int i = offset; i <= people.size() - k; ++i) {
combination.push_back(people[i]);
go(i + 1, k - 1);
combination.pop_back();
}
}
vector<vector<int>> combinationSum3(int k, int n) {
for (int i = 1; i <= 9; i++) { people.push_back(i); }
go(0, k);
vector<vector<int>> comb3;
// check if == sum
for (auto i : final_combs) {
int sum = accumulate(i.begin(), i.end(), 0);
if (sum == n) {
comb3.push_back(i);
}
}
return comb3;
}