How to consider only first two elements from *pointer - c++

Question

From the below code you can see that the vector array has the same number twice or more than. What I want to do is to find the first two same number's position from the pointer *ptr
#include<iostream>
#include<iterator> // for iterators
#include<vector> // for vectors
using namespace std;
int main()
{
vector<int> ar = { 1,8,2, 2, 2, 5,7,7,7,7,8 };
// Declaring iterator to a vector
vector<int>::iterator ptr;
// Displaying vector elements using begin() and end()
cout << "The vector elements are : ";
for (ptr = ar.begin(); ptr < ar.end(); ptr++)
cout << *ptr << " ";
return 0;
}
Let's assume I want to print out the first two position and elements of 7 by dereferencing the pointer *ptr. Should I use an if a condition like
int *array = ptr.data();
for( int i =0; i < ar.size(); i++) {
if( array[i] - array[i+1]+ ==0)
cout<<array[i]<<endl;
}
But how would I guarantee that it is not looking for the only first two same elements from *ptr?
UPDATE
Clearing the question:
The reason I always want to know the first and second position of the same element from the dereferencing the pointer is that later I will do some study, and in that study, I will be given some time associated with the first and second position of the same number. The problem, I wanted to ignore the same elements which are still repetitive after the second time is because I want to ignore these element positions in my calculations.
For example, if you print out the code you would find the element: **The vector elements are 1 8 2 2 2 5 7 7 7 7 8 **. In this case, the first two positions of the element 2, is [2], and [3], therefore I would like to ignore the position [4]. Another thing to mention, that I don't care if the value or consequent or not[I mean for example 828, or 888, I would consider both]. For example, the number 8 is in location array[1], and in the [10]. I would also consider this.

Answer 1

Create a map where each value is stored as key, mapped to a list of indices:
std::unordered_map<int, std::vector<size_t>> indexMap;
Loop over your initial values and fill the map:
for (size_t index = 0; index < ar.size(); index++)
{
indexMap[ar[index]].push_back(index);
}
Now you can loop over your map and work with every value that has 2 or more indices and only use the first 2 indices for whatever you want to do:
for (auto const& [value, indices] : indexMap)
{
if (indices.size() < 2)
continue;
size_t firstIndex = indices[0];
size_t secondIndex = indices[1];
// do whatever
}
(If you don't use C++17 or up, use for (auto const& pair : indexMap), where pair.first is value and pair.second is indices.)

Answer 2

You could use map or unordered_map to register indexes of each value.
Here's a simple demo of the concept:
#include<iostream>
#include<vector>
#include<map>
using namespace std;
int main() {
vector<int> ar{ 1, 8, 2, 2, 2, 5, 7, 7, 7, 7, 8 };
map<int, vector<size_t> > occurrences{ };
for (size_t i = 0; i < ar.size(); ++i) {
occurrences[ar[i]].push_back(i);
}
for (const auto& occurrence:occurrences) {
cout << occurrence.first << ": ";
for (auto index: occurrence.second) {
cout << index << " ";
}
cout << endl;
}
return 0;
}
Output:
1: 0
2: 2 3 4
5: 5
7: 6 7 8 9
8: 1 10