Generate Partition of integer into k different integers (C++) - c++

Question

I have problem with the constraint
I have to partition it to "exact" k "different" integers
e.g. 10 = 1+2+7 -> valid
10 = 2+2+6 -> invalid
I don't want to print them, I want to store them in to vectors or something
I am thinking of recursive solution, but still can't come up with a efficient way to store them...(the solution I found can only print it)
And I think it should store in vector<vector>
the struct should be like this?
Partition(){
....
....
}
Partition_main(){
...
Partition()
....
}

Answer 1

In practice, the problem is to enumerate all the solutions.
This can be done by a DFS.
In the following code, the DFS is implemented with a help of a FIFO (i.e. a std::queue).
The little trick is, for a given candidate at a given step, to calculate the minimum
sum that can be obtained in the next steps.
If this minimum sum is larger than n, then we can stop the research in this direction.
#include <iostream>
#include <vector>
#include <queue>
#include <cassert>
struct Parti {
std::vector<int> v;
int sum = 0;
};
std::vector<std::vector<int>> get_partitions (int n, int k) {
std::vector<std::vector<int>> result;
std::queue<Parti> fifo;
Parti start;
fifo.push(start);
while (!fifo.empty()) {
Parti elt = fifo.front();
fifo.pop();
int sum = elt.sum;
int remain = k-elt.v.size();
if (remain == 1) {
int last = n - sum;
if (k > 1) assert (last > elt.v.back());
elt.v.push_back(last);
result.push_back (elt.v);
continue;
}
int i;
if (elt.v.size() == 0) i = 1;
else i = elt.v.back() + 1;
while (true) {
int min_sum = sum + remain*(2*i + remain - 1)/2;
if (min_sum > n) break;
Parti new_elt = elt;
new_elt.v.push_back(i);
new_elt.sum += i;
fifo.push (new_elt);
++i;
};
}
return result;
}
int main() {
int n = 15;
int k = 4;
auto res = get_partitions (n, k);
if (res.size() == 0) {
std::cout << "no partition possible\n";
return 0;
}
for (const auto& v: res) {
std::cout << n << " = ";
for (int i = 0; i < v.size(); ++i) {
std::cout << v[i];
if (i == v.size()-1) {
std::cout << "\n";
} else {
std::cout << " + ";
}
}
}
return 0;
}