Inplace unified vectors

Question

I need an efficient method to combine inplace of a sorted vector with another sorted vector. In inplace, I mean that the algorithm should not create a whole new vector or other storage to store the union, even temporarily. Instead, the first vector should simply grow precisely in the number of new elements.

Sort of:

void inplace_union(vector & A, const vector & B);

Where after A contains all the elements of A union B and is sorted.

std::set_unionin <algorithm>does not work, because it overwrites your destination to be A .

In addition, can this be done in just one pass over two vectors?

Edit: Items that are in both A and B should only appear once in A.

Answer 1

I believe you can use an algorithm std::inplace_merge. Here is a sample code:

void inplace_union(std::vector<int>& a, const std::vector<int>& b)
{
    int mid = a.size(); //Store the end of first sorted range

    //First copy the second sorted range into the destination vector
    std::copy(b.begin(), b.end(), std::back_inserter(a));

    //Then perform the in place merge on the two sub-sorted ranges.
    std::inplace_merge(a.begin(), a.begin() + mid, a.end());

    //Remove duplicate elements from the sorted vector
    a.erase(std::unique(a.begin(), a.end()), a.end());
}

Answer 2

Yes, this can be done in place, but in O (n) time, assuming that both inputs are sorted, and with one pass along both vectors. Here's how:

Extend A(target vector) to B.size()- make room for our new elements.

, B A. → ( ), , , A. , B B. .

:

A: [ 1, 2, 4, 9 ]
B: [ 3, 7, 11 ]

* = iterator, ^ = where we're inserting, _ = unitialized
A: [ 1, 3, 4, 9*, _, _, _^ ]   B: [ 3, 7, 11* ]
A: [ 1, 3, 4, 9*, _, _^, 11 ]  B: [ 3, 7*, 11 ]
A: [ 1, 3, 4*, 9, _^, 9, 11 ]  B: [ 3, 7*, 11 ]
A: [ 1, 3, 4*, 9^, 7, 9, 11 ]  B: [ 3*, 7, 11 ]
A: [ 1, 3*, 4^, 4, 7, 9, 11 ]  B: [ 3*, 7, 11 ]
A: [ 1, 3*^, 3, 4, 7, 9, 11 ]  B: [ 3, 7, 11 ]

-: std::inplace_merge? ( , , ?)

Answer 3

set_difference , , , .

, set_difference , , , , .

Note: this means that we will iterate over the source code twice.

#include <algorithm>
#include <boost/function_output_iterator.hpp>

// for the test
#include <vector>
#include <iostream>


struct output_counter
{
   output_counter(size_t & r) : result(r) {}
   template <class T> void operator()(const T & x) const { ++result; }
private:
   size_t & result;
};


// Target is assumed to work the same as a vector
// Both target and source must be sorted
template <class Target, class It>
void inplace_union( Target & target, It src_begin, It src_end )
{
   const size_t mid = target.size(); // Store the end of first sorted range

   // first, count how many items we will copy
   size_t extra = 0;
   std::set_difference( 
         src_begin, src_end,
         target.begin(), target.end(),
         boost::make_function_output_iterator(output_counter(extra)));

   if (extra > 0) // don't waste time if nothing to do
   {
      // reserve the exact memory we will require
      target.reserve( target.size() + extra );

      // Copy the items from the source that are missing in the destination
      std::set_difference( 
            src_begin, src_end,
            target.begin(), target.end(),
            std::back_inserter(target) );

      // Then perform the in place merge on the two sub-sorted ranges.
      std::inplace_merge( target.begin(), target.begin() + mid, target.end() );
   }
}



int main()
{
   std::vector<int> a(3), b(3);
   a[0] = 1;
   a[1] = 3;
   a[2] = 5;

   b[0] = 4;
   b[1] = 5;
   b[2] = 6;

   inplace_union(a, b.begin(), b.end());

   for (size_t i = 0; i != a.size(); ++i)
      std::cout << a[i] << ", ";
   std::cout << std::endl;

   return 0;
}

Compiled with gain headers, result:

$ ./test 
1, 3, 4, 5, 6,