Sort one vector using another in nonincreasing order

Question

This mean that while I am sorting the v2 in the nonincreasing order, the v1 should looks like this:

Vectors look as following.

v1  = {0, 5, 5, 2,  10};
v2  = {0 ,2, 6, 20, 5};

The output:

v1 = {2,  5, 10, 5, 0};
v2 = {20, 6,  5, 2, 0};

I was trying to solve that problem mixing std::sort and lambdas. That is why I have read several questions about std::sort, there were no answers which solve the problem similar to mine so that is why I am asking.

I am especially interested in the answers which contains it's usage or other features of C++11 and C++14.

It is not a question like:

"I completely do not know what to do."

I know how to achieve the output using C++98, but I am wondering if there is an more efficient and prettier way to do it.

Big thanks for your help :)

Answer 1

You could zip, sort, and unzip.

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

//converts two vectors into vector of pairs
template <typename T, typename U>
auto zip(T t, U u) {
  std::vector<std::pair<typename T::value_type,typename U::value_type>> pairs;
  for (size_t i = 0; i < t.size(); ++i){
    pairs.emplace_back(u[i],t[i]);
  }
  return pairs;
}

//converts a vector of pairs, back into two two vectors
template <typename T, typename U, typename V>
void unzip(V pairs, T & t, U & u) {
  for (auto const& it: pairs){
    u.emplace_back(it.first);
    t.emplace_back(it.second);
  }
}


int main(){

  //vectors
  std::vector<int> v1  = {0, 5, 5, 2,  10};
  std::vector<int> v2  = {0 ,2, 6, 20, 5};

  //zip vectors
  auto pairs = zip(v1,v2);

  //sort them
  std::sort(pairs.begin(),pairs.end(),std::greater<>());

  //unzip them
  v1.clear();
  v2.clear();
  unzip(pairs,v1,v2);

  //print
  std::cout << '\n';
  for (auto i: v1) std::cout << i << ' ';
  std::cout << '\n';
  for (auto i: v2) std::cout << i << ' ';
  std::cout << '\n';

}