How do I compare an element of array in c/c++? - c++

Question

int arr[] = {10, 20, 20, 10, 10, 30, 50, 10, 20};
I want to compare each element in an array so that I can make pairs of similar numbers using C/C++.
In above example there are three pairs (10-10, 20-20, 10-10).
I want to find number of pairs in given array(i.e. 3).
Can anyone please help me with logic?

Answer 1

The approach using C can differ from the approach using C++ because for example in C++ you can use standard containers and algorithms while in C they are absent.
So I will show a solution that can be implemented in the both languages.
Here you are. The program is written using C but it is easy can be transformed in a C++ program by substituting the header and only one statement: the output statement.
#include <stdio.h>
int main( void )
{
int a[] = { 10, 20, 20, 10, 10, 30, 50, 10, 20 };
const size_t N = sizeof( a ) / sizeof( *a );
size_t total = 0;
for ( size_t i = 1; i < N; i++ )
{
size_t count = 1;
for ( size_t j = 0; j < i; j++ )
{
if ( a[i] == a[j] ) ++count;
}
if ( count % 2 == 0 ) ++total;
}
printf( "The number of pairs of equal elements is %zu\n", total );
return 0;
}
The program output is
The number of pairs of equal elements is 3
In C++ you can use for example the following alternative approach using the standard container std::map or std::unordered_map.
#include <iostream>
#include <map>
int main()
{
int a[] = { 10, 20, 20, 10, 10, 30, 50, 10, 20 };
size_t total = 0;
std::map<int, size_t> m;
for ( const auto &item : a ) ++m[item];
for ( const auto &item : m ) total += item.second / 2;
std::cout << "The number of pairs of equal elements is " << total << '\n';
return 0;
}
The program output is the same as shown above.
The number of pairs of equal elements is 3

Answer 2

And here the C++ solution.
Same approach as Vlad. Simply count the groups of all values and divide this counter by 2, because we are lokking for pairs.
Then count the overall counts:
#include <iostream>
#include <map>
#include <algorithm>
#include <numeric>
int main() {
// The data
int arr[] = { 10, 20, 20, 10, 10, 30, 50, 10, 20 };
// The counter for occurences of one value in the array
std::map<int, size_t> counter{};
// Do Count
for (int i : arr) counter[i]++;
// Devide all counts by 2. Thats, because we are looking for pairs
std::for_each(counter.begin(), counter.end(), [](auto& p) { p.second /= 2;});
// Calculate the overall sum
size_t pairCounter = std::accumulate(counter.begin(), counter.end(), 0U,
[](const size_t v, const auto& p) { return v + p.second; });
std::cout << "\nNumber of pairs: " << pairCounter << "\n";
return 0;
}