How to reduce complexity of this code - c++

Question

Please can any one provide with a better algorithm then trying all the combinations for this problem.
Given an array A of N numbers, find the number of distinct pairs (i,
j) such that j >=i and A[i] = A[j].
First line of the input contains number of test cases T. Each test
case has two lines, first line is the number N, followed by a line
consisting of N integers which are the elements of array A.
For each test case print the number of distinct pairs.
Constraints:
1 <= T <= 10
1 <= N <= 10^6
-10^6 <= A[i] <= 10^6 for 0 <= i < N
I think that first sorting the array then finding frequency of every distinct integer and then adding nC2 of all the frequencies plus adding the length of the string at last. But unfortunately it gives wrong ans for some cases which are not known help. here is the implementation.
code:
#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
long fun(long a) //to find the aC2 for given a
{
if (a == 1) return 0;
return (a * (a - 1)) / 2;
}
int main()
{
long t, i, j, n, tmp = 0;
long long count;
long ar[1000000];
cin >> t;
while (t--)
{
cin >> n;
for (i = 0; i < n; i++)
{
cin >> ar[i];
}
count = 0;
sort(ar, ar + n);
for (i = 0; i < n - 1; i++)
{
if (ar[i] == ar[i + 1])
{
tmp++;
}
else
{
count += fun(tmp + 1);
tmp = 0;
}
}
if (tmp != 0)
{
count += fun(tmp + 1);
}
cout << count + n << "\n";
}
return 0;
}

Answer 1

Keep a count of how many times each number appears in an array. Then iterate over the result array and add the triangular number for each.
For example(from the source test case):
Input:
3
1 2 1
count array = {0, 2, 1} // no zeroes, two ones, one two
pairs = triangle(0) + triangle(2) + triangle(1)
pairs = 0 + 3 + 1
pairs = 4
Triangle numbers can be computed by (n * n + n) / 2, and the whole thing is O(n).
Edit:
First, there's no need to sort if you're counting frequency. I see what you did with sorting, but if you just keep a separate array of frequencies, it's easier. It takes more space, but since the elements and array length are both restrained to < 10^6, the max you'll need is an int[10^6]. This easily fits in the 256MB space requirements given in the challenge. (whoops, since elements can go negative, you'll need an array twice that size. still well under the limit, though)
For the n choose 2 part, the part you had wrong is that it's an n+1 choose 2 problem. Since you can pair each one by itself, you have to add one to n. I know you were adding n at the end, but it's not the same. The difference between tri(n) and tri(n+1) is not one, but n.