C++ please help I will give you a good rating!!!!! Implement the following function that returns a vector of vectors of Item, each of which is a subset of elements chosen from the given vector a[first…last]:  /* if a is {1,2,3}, first=0, last=2, the function shall returns a vector of the following vectors:    {}, {1}, {2}, {3}, {1,2},{1,3}, {2,3},{1,2,3}, all subsets of a[0…2].    Precondition: last-first+1>=1, i.e., there is at least one element in the a[first…last]    Note 1)if the length of a[first…last] is n, then the function should return a vector of 2n vectors      2) The order of these subsets does not need to match what’s listed here…  */ vector> subsets (const vector & a, int first, int last)

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C++ please help I will give you a good rating!!!!!

Implement the following function that returns a vector of vectors of Item, each of which is a subset of elements chosen from the given vector a[first…last]:

 /* if a is {1,2,3}, first=0, last=2, the function shall returns a vector of the following vectors:

   {}, {1}, {2}, {3}, {1,2},{1,3}, {2,3},{1,2,3}, all subsets of a[0…2].

   Precondition: last-first+1>=1, i.e., there is at least one element in the a[first…last]

   Note 1)if the length of a[first…last] is n, then the function should return a vector of 2n vectors

     2) The order of these subsets does not need to match what’s listed here… 

*/

vector<vector<int>> subsets (const vector<int> & a, int first, int last)

CODE:

#include <iostream>

#include <algorithm>

#include <vector>

using namespace std;

 

void PrintVector (const vector<int> & v){

cout <<"[";

for (auto e:v){

cout<<e<<" ";

}

cout <<"]";

 

}

 

/* check if we can use values in L[left...right] to make a sum of value, and find

the best solution, i.e., smallest set of coins tht make this value

 @param L, first, last: specify a sub-vector where coins/values are chosen from

 @param value: the sum/value we want to make

 @pre-condition: all parameters are initialized, L[] and value are non-negative

 @post-condition: return true/false depending on the checking result, if return true,

   used vector contains coins that make up the value, with the minimul # of elements from 

   L [first...last]

*/

bool CoinChange (vector<int> & L, int first, int last, int value, vector<int> & used)

{

 

    if (value==0)

    {

  used.clear();

 

return true;

    }

 

   if (first>last) //no more coins to use

   {

used.clear();

 

return false;

   }

 

   if (value<0)

   {

used.clear();

 

return false;

   }

 

   //general case below

 

   vector<int> used1;

   bool ret1= CoinChange (L, first, last-1, value-L[last], used1);

   if (ret1) 

        // used1 include all values from L[first...last-1] that add up to valeu-L[last]

        used1.push_back (L[last]);

        //now: used1 include all coins used from L[first...last} that add up to value

 

   vector<int> used2;

   // If not using L[last]... 

   bool ret2 = CoinChange (L, first, last-1, value, used2);

 

   if (ret1 && ret2) {

if (used1.size() > used2.size())

used = used2;

else

used = used1;

        return true;

   } else if (ret1) {

used = used1;

return true;

   } else if (ret2){

used = used2;

return true;

   } else {

used.clear();

return false;

   }

 

  

 

 

bool CoinChangeK (const vector<int> & coins, int first, int last, int value, int K)

{

   return true;

}

 

bool UnlimitedCoinChange (const vector<int> & coins, int value, vector<int>& bestSolution)

{

   

   return true;

}

 

int main()

{

   vector<int> coins{2,5,3,10};

   vector<int> used;

 

   vector<int> values{4, 6,15, 18, 30, 41}; //use this to test

   

 

   //This part demo the CoinChange function: optimization problem

/*

   for (auto v: values) {

   //Todo: replace CoinChange with your CoinChangeUnlimited function... 

   if (CoinChange (coins, 0, coins.size()-1, v, used))

   {

cout <<"value="<<v <<" True\n";

//display used vector

        for (int i=0;i<used.size();i++)

cout <<used[i]<<" ";

        cout<<endl;

   }

   else 

cout <<"Value=" << v<<" False"<<endl;

   }

*/

 

   //Test CoinChangeK

  cout <<"Enter coinchangek or unlimited to test the corresponding function:";

  string command;

 

  cin >> command;

 

  if (command=="coinchangek"){  

  //we cannot make 20 using 2 or fewer coins

  if (CoinChangeK (coins, 0, coins.size()-1, 20, 2)!=false || 

      CoinChangeK (coins, 0, coins.size()-1, 5, 1)!=true)

  {

cout <<"fail coinchangek tests\n";

       return 1; //faild coinchangeK test 

  }

else{

cout <<"pass coinchangek tests\n";

return 0; //pass coinchangeK test

}

  } else if (command=="unlimited"){

   //Test UnlimitedCoinChange 

vector<int> bestSolution;

 

   if (UnlimitedCoinChange (coins, 1,bestSolution)!=false) {

      cout <<"Failed UnlimitedCoinChange case 1\n";

return 1; //failed unlimited test 

}

 

   if (UnlimitedCoinChange (coins, 15, bestSolution)!=true) {

cout <<"Failed UnlimitedCoinChange case 2\n";

return 1;

}

vector<int> expectedSol{5,10}; 

        sort (bestSolution.begin(), bestSolution.end());

if (bestSolution!=expectedSol){

cout <<"Failed UnlimitedCoinChange case 2\n";

return 1;

 

}

 

 

   if (UnlimitedCoinChange (coins, 30, bestSolution)!=true) {

cout <<"Failed UnlimitedCoinChange case 3\n";

return 1;

}

vector<int> expectedSol3{10,10,10}; 

        sort (bestSolution.begin(), bestSolution.end());

if (bestSolution!=expectedSol3){

cout <<"Failed UnlimitedCoinChange case 3\n";

return 1;

 

}

 

        cout <<"Pass unlimitedCoinChange cases\n";

        return 0;

  }

 

}

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