HackerEarth Shubham and Xor problem solution

In this HackerEarth Shubham and Xor problem solution You are given an array of n integer numbers a1, a2, .. ,an. Calculate the number of pair of indices (i,j) such that 1 <= i < j <= n and ai xor aj = 0.

HackerEarth Shubham and Xor problem solution.

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define inf 1000000000000
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define S second
#define F first
#define boost1 ios::sync_with_stdio(false);
#define boost2 cin.tie(0);
#define mem(a,val) memset(a,val,sizeof a)
#define endl "\n"
#define maxn 1000005

ll arr[maxn];

inline ll c2(ll x)
{
    return (x*(x-1))/2; 
}
int main()
{
    boost1;boost2;
    ll i,j,n,x,y,ans=0,ptr;
    cin>>n;
    for(i=1;i<=n;i++)
    cin>>arr[i];
    sort(arr+1,arr+n+1);
    ptr=1;
    for(i=2;i<=n;i++)
    {
        if(arr[i]==arr[i-1])
        continue;
        x=(i-1)-ptr+1;
        ans+=c2(x);
        ptr=i;
    }
    x=n-ptr+1;
    ans+=c2(x);
    cout<<ans;
    return 0;
}

Second solution

#include <bits/stdc++.h>
using namespace std;

int main(){
    int i,j,n,a;
    long long ans = 0;
    unordered_map<int, int> mp;
    scanf("%d", &n);
    for(i = 0; i < n; i++){
        scanf("%d", &a);
        mp[a]++;
        ans -= (1ll) * (mp[a] - 1) * (mp[a] - 2) / 2;
        ans += (1ll) * (mp[a]) * (mp[a] - 1) / 2;
    }
    printf("%lld\n", ans);
    return 0;
}