C++ arrays sort - c++

Question

I have to solve a problem that consist of sorting a list of number in c++.
The condition is :
the positive, and 0 must be sorted ascending
negative numbers- descending
if A = {4,-8,7,-6,0,-7,5} the at the final B = {0,4,5,7,-6,-7,-8 }
#include <iostream>
#include <cstring>
using namespace std;
int main()
{
int i,j,n,A[100],B[100],A_nenegative[100],A_negative[100],aux,m=0,k=0;
cout<<"max elements"; cin>>n;
cout<<"elements are"<<endl;
for(i=0;i<n;i++)
{
cin>>A[i];
}
for(i=0;i<n;i++)
for(j=i+1;j<n;j++)
if(A[i]>A[j])
{
aux=A[i];
A[i]=A[j];
A[j]=aux;
}
for(i = 0; i< n; i++)
if(A[i]>=0) {
A_nenegative[i]=A[i];
B[i]=A_nenegative[i];
k++;
}
for(i=0;i<n;i++)
for(j=i+1;j<n;j++)
if(A[i]<A[j])
{
aux=A[i];
A[i]=A[j];
A[j]=aux;
}
for(i=0;i<n;i++)
if(A[i]<0)
{
A_negative[i]=A[i];
m++;
}
}
here is where i stopped. I sort positive numbers in A_nenegative and negative numbers in A_negative. So the question is how can attribute to B first- A_nenegative and then A_negative.
I try after sorting positive numbers immediate to attribute to B , something like that:
for(i = 0; i< n; i++)
if(A[i]>=0) {
A_nenegative[i]=A[i];
B[i]=A_nenegative[i];
k++;
}
But i don't know what to do next

Answer 1

Just use predicate.
std::sort(std::begin(a), std::end(a), [](int a, int b) {
if(a >= 0) {
if(b >= 0) return a < b;
else return true;
}
else {
if(b >= 0) return false;
else return a > b;
}
});
Predicate returns true if a needs to go before b in array, false - otherwise.
So, if a >= 0 and b >= 0 --- we check whether a < b (sort non-negative values in ascending order).
if a >= 0 and b < 0 --- then b needs to go after a, because all negative values go after non-negative.
if a < 0 and b >= 0 --- then b needs to go before a.
if a < 0 and b < 0 --- check whether a > b (to sort in descending order).

Answer 2

Use appropriate predicate with std::sort:
std::sort(std::begin(A), std::end(A), [](int lhs, int rhs) {
auto score = [](int n){ return std::make_tuple(n < 0, std::abs(n)); }
return score(lhs) < score(rhs);
});

Answer 3

#include <iostream>
#include <cstring>
using namespace std;
int main()
{
int n,A[100],B[100],Merge_Array[200],aux;
cout<<"max elements"; cin>>n;
cout<<"elements are"<<endl;
int temp,index_A=0,index_B=0;
for(int i=0;i<n;i++)
{
cin>>temp;
if(temp>=0){
A[index_A++]=temp; //One array for positive
}
else{
B[index_B++]=temp; //Another one for negative
}
}
//Sort positive in ascending
for(int i=0;i<index_A;i++)
{
for(int j=i+1;j<index_A;j++)
if(A[i]>A[j])
{
aux=A[i];
A[i]=A[j];
A[j]=aux;
}
}
//Sort negative in descending
for(int i=0;i<index_B;i++)
for(int j=i+1;j<index_B;j++)
if(B[i]<B[j])
{
aux=B[i];
B[i]=B[j];
B[j]=aux;
}
//Merge A and B
for(int i=0;i<index_A;i++){
Merge_Array[i]=A[i];
}
for(int i=index_A,k=0;i<n;i++,k++){
Merge_Array[i]=B[k];
}
//Print the Merged Array
for(int i=0;i<n;i++){
cout<<Merge_Array[i];
}
}
Hi Cristyan. All others have answered your question. But I still see you are following C type coding for C++. In C++ you are free to declare your variables anywhere not necessarily at the start of the function.
Eg:- for(int i=0;;) is feasible.
Moreover, you have not used dynamic arrays. I know for starters using new and delete can be intimidating. But instead, you can use vectors which offers various functionalities, so pointing you to that vectors.

Answer 4

I like expressive C++:
#include <array>
#include <iostream>
int main() {
auto a = std::array<int, 7>{4, -8, 7, -6, 0, -7, 5};
auto it = std::partition(std::begin(a), std::end(a),
[](const int val) noexcept { return val >= 0; });
std::sort(begin(a), it);
std::sort(it, std::end(a), std::greater<int>());
for(const auto& elem : a) {
std::cout << elem << "\n";
}
}

Answer 5

I would iterate over your input and put every value >= 0 into an array POS and any value < 0 into an array NEG.
Then sort both array ascending and separately and use the absolute value to compare the elements (no effect in POS but required for NEG).
For last push NEG to POS and you have it.
Think thats simple and straight forward.