c++,partitioning , Minimally partitioning a vector of objects (C++)

## Question:

Tag: c++,partitioning

I have an `std::vector` of objects where each element in the vector more or less looks like:

``````struct Obj {
int group;
};
``````

The entries in the vector are in no particular order. Ordinarily, when partitioning, one might want to typically group elements in the same partition that have something in common, however, what I am wanting to in my case is actually rearrange the entries in this vector and partition it in such a way as to use the absolute minimum number of partitions possible where each element in a single partition belongs to a different group from each other element in the same partition.

Is this possible to do without iterating over every single permutation of the vector and seeing how many partitions each permutation has?

edit:

An example was asked for, so I'll attempt to provide one.

If the initial vector of objects was

``````[ {1}, {2}, {3}, {2}, {3}, {1}, {4}, {5}, {3}, {2} ]
``````

The optimal partitioning would be to divide it into three partitions, as follows:

``````[ {1}, {2}, {3}, {4}, {5} ] [ {1}, {2}, {3} ] [{2}, {3} ]
``````

So that within each partition, all of entries belong to a different group.

If I understand your requirements correctly, then the “minimum number of partitions” is given by the maximum frequency of a single value in the original vector. So you can create a histogram and then find the maximum entry in it. (This is linear in the size of the vector.) Now create m vectors (where m is the maximum frequency as just determined) and assign each of the m identical values to one of them. It is guaranteed that you can distribute the remaining elements in a way such that no duplicates occur in the partitions.

In pseudo-code for input-vector v of size n:

• initialize an empty histogram H
• for each item x in v:
• increment H [ x ]  by one, zero-initializing it before if no such bin is already present
• m ← maximum frequency in H
• initialize empty vectors v1, …, vm
• for each value x were H [ x ]  ≥ 0:
• for i ← 1 to H [ x ] :
• append x to vi

Note that this works fine if the objects in your vector have the key that determines whether they are equal as their only data member. However, if they have more state that needs to be preserved but does not participate in determining equality, the procedure is easily adjusted to account for that.

• initialize an empty histogram H
• for each item x in v:
• increment H [ key(x) ]  by one, zero-initializing it before if no such bin is already present
• m ← maximum frequency in H
• initialize empty vectors v1, …, vm
• for each value x in v:
• iH [ key(x) ]
• append x to vi
• decrement H [ key(x) ]  by one

If you want a quick solution, you can use a `std::unordered_map<int, int>` for your histogram.

Here is how an (eventually somewhat over-generalized) implementation could look like in C++14.

``````#include <algorithm>            // std::max_element
#include <functional>           // std::hash, std::equal_to
#include <iterator>             // std::iterator_traits
#include <unordered_map>        // std::unordered_map
#include <vector>               // std::vector

template<typename FwdIterT,
typename ValueT = typename std::iterator_traits<FwdIterT>::value_type,
typename ValueHashT = std::hash<ValueT>,
typename ValueEqCmpT = std::equal_to<ValueT>>
decltype(auto)
min_partition(const FwdIterT begin, const FwdIterT end)
{
std::vector<std::vector<ValueT>> partitions {};
std::unordered_map<ValueT, int, ValueHashT, ValueEqCmpT> histo {};
for (auto iter = begin; iter != end; ++iter)
histo[*iter]++;
const auto cmpfreq = [](const auto& lhs, const auto& rhs){
return lhs.second < rhs.second;
};
const auto maxbin = std::max_element(histo.cbegin(), histo.cend(), cmpfreq);
partitions.resize(maxbin->second);
for (auto iter = begin; iter != end; ++iter)
partitions.at(histo.at(*iter)-- - 1).push_back(*iter);
return partitions;
}
``````

It can be used like this.

``````#include <iostream>             // std::cout
#include <string>               // std::string
#include <utility>              // std::begin, std::end

int
main(int argc, char * * argv)
{
using std::begin;
using std::end;
for (int i = 1; i < argc; ++i)
{
const std::string text {argv[i]};
const auto partitions = min_partition(begin(text), end(text));
std::cout << "input:  " << text << "\n";
std::cout << "output: " << partitions.size() << " partitions\n\n";
for (auto it1 = begin(partitions); it1 != end(partitions); ++it1)
{
std::cout << "[";
for (auto it2 = begin(*it1); it2 != end(*it1); ++it2)
std::cout << (it2 != begin(*it1) ? ", " : "") << *it2;
std::cout << "]\n";
}
if (i != argc - 1)
std::cout << "\n\n";
}
}
``````

If given some well-known strings as input, it produces the following output.

``````input:  WEWEREARRESTEDAFTERDADATEDEEREGGS
output: 10 partitions

[W, F, A, T, D, R, E, G, S]
[W, S, T, R, A, D, E, G]
[R, T, A, D, E]
[A, R, D, E]
[R, E]
[E]
[E]
[E]
[E]
[E]

output: 8 partitions

[H, G, S, L, A, D]
[D, L, S, A]
[L, D, A, S]
[S, D, A]
[A]
[A]
[A]
[A]

input:  THEQUICKBROWNFOXJUMPSOVERTHESLEAZYDOG
output: 4 partitions

[Q, I, C, K, B, W, N, F, X, J, U, M, P, V, R, T, H, S, L, E, A, Z, Y, D, O, G]
[T, H, U, R, S, O, E]
[O, E]
[E, O]
``````

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