HackerEarth Shubham and Xor problem solution YASH PAL, 31 July 202415 February 2026 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 1000005ll 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("%lldn", ans); return 0;} coding problems solutions HackerEarth HackerEarth