sorting an array and maintaining element's old index - c++

Question

I have an array A:
A = [10 11 3 15 8 7]
index = 0 1 2 3 4 5
I want to sort this array.After sorting I want the information of old index.For this I can create a structure like this.
struct VnI{
int value;
int index;
};
sorting the array of structure with respect to value solve my problem.But I want to know that is it possible to solve this using sort or any other function in C++11.
I have tried this way:
struct VnI{
int V;
int I;
};
bool comparator(VnI x,VnI y){
if(x.V < y.V)
return true;
return false;
}
int maximumGap(const vector<int> &A) {
vector<VnI> B;
for(int i = 0;i < A.size();i++){
B[i].I = i;
B[i].V = A[i];
}
sort(B.begin(),B.end(),comparator);
for(int i = 0;i < B.size();i++){
cout<<B[i].I<<" "<<B[i].V<<endl;
}
}
But I got runtime error.
Please help.

Answer 1

This code is wrong:
vector<VnI> B;
for(int i = 0;i < A.size();i++){
B[i].I = i;
B[i].V = A[i];
}
When you write B[i], it assumes that B is at least of size i+1. Since the maximum value of i (which you used an index to B) is A.size()-1. The assumption in your code is that B is at least of size A.size(). This assumption is wrong — the fact is that B is of size 0.
Unfortunately operator[] of std::vector doesn't check for out of range index. If you use at(), the code will throw std::out_of_range exception:
vector<VnI> B;
for(int i = 0;i < A.size();i++){
B.at(i).I = i;
B.at(i).V = A[i];
}
Now this would throw std::out_of_range exception.
Anyway, one simple fix could be this:
vector<VnI> B (A.size()); //initialize B with the size of A.
for(int i = 0;i < A.size();i++){
B[i].I = i;
B[i].V = A[i];
}
However, I'd suggest this solution:
vector<VnI> B;
B.reserve(A.size());
for(int i = 0;i < A.size(); i++){
B.emplace_back(i, A[i]);
}
I'd also suggest you read more about std::vector, especially the following functions:
size()
capacity()
resize()
reserve()
push_back()
operator[]
at()
emplace_back()
and all the constructors.
Also, learn to naming your variables properly and be consistent with it.
Hope that helps.

Answer 2

do you pefer to use vector and pair?
each pair has "first" and "second", put "first"=value to sort,"second"=original index, create a pair for each element and put them into vector to sort:
int N[]={10,11,3,15,8,7};
std::vector<std::pair<int,int> > v;
//create pair for each element
for(int i=0;i<sizeof(N)/sizeof(int);i++){
//first is value of array,second is original index
v.push_back(std::make_pair(N[i],i));
}
//sort the vector of pair
sort(v.begin(),v.end());
//get original index from second of pair
for(std::pair<int,int>& p : v){
std::cout << p.first << ":" << p.second << std::endl;
}
output
3:2
7:5
8:4
10:0
11:1
15:3

Answer 3

Normally what is done is the opposite... i.e. given an array x of elements compute an array of integers ix so that x[ix[i]] appears to be sorted under a certain criteria.
This allows representing the container with different orderings without actually having to move/copy the elements.
With C++11 this can easily be done using lambdas:
// Build the index vector ix
std::vector<int> ix(x.size());
for (int i=0,n=x.size(); i<n; i++) ix[i] = i;
// Sort ix according to the corresponding values in x
// (without touching x)
std::sort(ix.begin(), ix.end(),
[&x](int a, int b) { return x[a] < x[b]; });
This ix index array is what you are asking for (i.e. the "old" position of an element: ix[i] is where the i-th element of the sorted list was in the original array) and there is no need to modify the input array.

Answer 4

You are trying to sort a list of custom objects, answerd here:
SO Link
Once you list the list of VnI objects you can then access there old index's through the I member that I presume is the index.