How to efficiently remove elements from vector C ++

The most efficient way to remove an element from a vector is the writeback trick, but this only applies if you don't need ordering.

#include <vector>
#include <iostream>

int main()
{
    std::vector<int> v { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    auto it = v.begin() + 5;
    // replace the current element with the back of the vector,
    // then shrink the size of the vector by 1.
    *it = std::move(v.back());
    v.pop_back();

    for (auto n : v) {
        std::cout << n << " ";
    }
    std::cout << "\n";
}

      

        
        
        
      

    

http://ideone.com/0jbWHZ

If you know there will be many deletions or a very large vector, you can save efficiency by using this trick, remembering to keep ++ your current iterator after deletion, and std::sort()

vector when you are through.

--- Edit ---

#include <algorithm>
#include <iostream>
#include <vector>

//! Efficiently remove an element from a vector without
//! preserving order. If the element is not the last element
//! in the vector, transfer the last element into its position
//! using a move if possible.
//! Regardless, we then shrink the size of the vector deleting
//! the element at the end, which will either be destructed or
//! the element we were deleting.
//! @note: Effectively invalidates the current iterator.
template<class ValueType>
bool unstable_remove(
    typename std::vector<ValueType>& container,
    typename std::vector<ValueType>::iterator it
    )
{
    // Leave in-situ if we are already the tail element.
    auto lastEl = container.end() - 1;
    if (it != lastEl) {
        // overwrite this element with what is in the last,
        // which should have the same effect as deleting this.
        *it = std::move(*lastEl);
    }
    // release the last cell of the vector, because it should
    // now either be destructed or contain the value we were
    // deleting.
    container.pop_back();
}

int main()
{
    std::vector<int> ints { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    auto it = ints.begin();
    while (it != ints.end()) {
        if ((*it % 3) == 0) {
            unstable_remove(ints, it);
            // do not pass go / ++it
            continue;
        }
        ++it;
    }
    std::cout << "after removes:\n";
    for (auto val : ints)
        std::cout << val << " ";
    std::cout << "\n";
    std::sort(ints.begin(), ints.end());
    std::cout << "after sort:\n";
    for (auto val : ints)
        std::cout << val << " ";
    std::cout << "\n";
}

      

        
        
        
      

    

Produces ( http://ideone.com/hGZPOC )

after removes:
1 2 10 4 5 8 
after sort:
1 2 4 5 8 10 

      

        
        
        
      

    

--- Edit 2 ---

Here's a cleanup of your code for readability, I've also dropped your trailing captures because ... you are removing elements.

#include <vector>
#include <cstdint>

using vec_t = std::vector<uint32_t>;
using vecpair_t = std::pair<vec_t, vec_t>;
using pairvec_t = std::vector<vecpair_t>;

int main(int argc, char** argv) {
    pairvec_t pairV1V2;
    for(auto itm = pairV1V2.begin(); itm != pairV1V2.end(); ++itm)
    {
        //Outer values
        auto& outerV1 = itm->first; // NOTE '&' - reference not copy!
        auto& outerV2 = itm->second;
        sort(outerV2.begin(), outerV2.end());
        for(auto itm2 = itm + 1; itm2 != pairV1V2.end(); ++itm2)
        {
            auto& innerV1 = itm2->first;
            auto& innerV2 = itm2->second;
            vec_t setDiffV1;

      

        
        
        
      

    

As for another way to optimize this - since your lists are sorted - go through both lists at the same time, comparing the values.

template<typename ValueType>
void dedupe_vectors(
    typename std::vector<ValueType>& lhs,
    typename std::vector<ValueType>& rhs
    )
{
    auto lit = lhs.begin();
    auto rit = rhs.begin();
    while (rit != rhs.end) {
        while (lit != lhs.end() && *lit < *rit)
            ++lit;
        if (lit == lhs.end())
            break;
        if (*lit == *rit) {
            v2.erase(rit);
            continue;
        }  
        ++rit;
    }
}

      

        
        
        
      

    

I know we are testing lit

vs twice lhs.end

. Take a look at the code your compiler generates with -O3 and see if it found it itself. If so, then you might be concerned about optimization.