How to count duplicate elements in array C++?

All Unordered_set operations are O(1) and O(n) in worse case, but usually remain constant (https://www.geeksforgeeks.org/unordered_set-in-cpp-stl/). In your case you are iterating twice your array, so it would be O(n2). In this case you iterate your array once O(n) and use O(1) set operations, so it would be a faster solution:

#include <iostream>
#include <bits/stdc++.h>

using namespace std;

int main()
{

    int arr[10] = {1, 9, 7, 8, 9, 6, 9, 8, 7, 1};
    unordered_set<int> numbersSet;
    unordered_set<int> duplicateSet;
    int duplicatesCount = 0;

    for (size_t i = 0; i < 10; i++)
    {
        int number = arr[i];
        numbersSet.insert(number);
        if (numbersSet.find(number) != numbersSet.end())
        {
            duplicatesCount++;
            duplicateSet.insert(number);
        }
    }

    cout << "\nAll elements duplicated : ";
    unordered_set<int>::iterator itr;
    for (itr = duplicateSet.begin(); itr != duplicateSet.end(); itr++)
        cout << (*itr) << endl;

    return 0;
}

All elements duplicated : 6
8
7
9
1

int main()
{
    const int SIZE = 10;
    int arr[SIZE] = {1, 9, 7, 8, 9, 6, 9, 8, 7, 1};
    int find[SIZE] = {0};
    int repeat[SIZE] = {0};
    int temp;
    for (int i = 0; i < SIZE; i++)
    {
        temp = arr[i];
        for (int j = 0; j < SIZE; j++)
        {
            if (i == j)
                continue;
            else{
                if (temp == arr[j])
                {
                    find[i] = temp;
                    repeat[j]++;
                }   
            }
        }       
    }
    for (int i = 0; i < SIZE; i++)
        cout << arr[i] << ' ';
    cout << endl;
    for (int i = 0; i < SIZE; i++)
        cout << find[i] << ' ';
    cout << endl;
    for (int i = 0; i < SIZE; i++)
        cout << repeat[i] + 1 << ' ';
    system("pause");
    return 0;
}