# HackerEarth Determining numbers problem solution

In this HackerEarth Determining numbers problem solution you are given an array of integers. The special property of the array is that exactly two different elements occur once while other elements occur twice. you are required to determine those two elements.

## HackerEarth Determining numbers problem solution.

`#include<bits/stdc++.h>#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>#define int long long#define double long double#define mod 1000000007#define maxN 1000005#define pb push_back#define mp make_pair#define mset(dp,val) memset(dp,val,sizeof dp);#define pii pair<int,int>#define pip pair<int,pii>#define vi vector<int>#define vpi vector<pii >#define endl "\n"#define fastIO ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)#define FOR(a,b,c) for(int(a) = b;a<=c;a++)#define repr(a,b,c) for(int(a) = b;a>=c;a--)#define rep(i,n) for(int(i) = 0;i<n;i++)#define fir first#define sec second#define beg begin()#define e end()#define len length()using namespace std;using namespace __gnu_pbds;typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;typedef tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;const int inf = 0x3f3f3f3f3f3f3f3fll;int modExpo(int x,int y){  int res = 1;  while(y > 0){    if(y&1)      res = (res*x)%mod;    y = y>>1;    x = (x*x)%mod;  }  return res;}int modInverse(int a,int m){  int m1 = m;  int y = 0,x = 1;  if(m == 1)return 0;  while(a > 1){    int q = a/m;    int t = m;    m = a%m,a = t;    t = y;    y = x - q*y;    x = t;  }  if(x < 0)x+=m1;  return x;}int arr[maxN];int32_t main(){  fastIO;  int n;  cin>>n;  rep(i,n)cin>>arr[i];  int xo = 0;  rep(i,n)xo^=arr[i];  int lo = 0;  while(true){    if((1ll<<lo)&xo)break;    lo++;  }  int tester = 1ll<<lo;  int setBit = 0,unSetBit = 0;  rep(i,n){    if(tester&arr[i])setBit^=arr[i];    else unSetBit^=arr[i];  }  if(setBit > unSetBit)swap(setBit,unSetBit);  cout<<setBit<<' '<<unSetBit;  return 0;}`

### Second solution

`#include <bits/stdc++.h>using namespace std;typedef long long ll; const int maxn = 1e6 + 14;int n;bool mark[maxn];int main(){    ios::sync_with_stdio(0), cin.tie(0);    cin >> n;    for(int i = 0; i < n; i++){        int x;        cin >> x;        mark[x] ^= 1;    }    for(int i = 1; i < maxn; i++)        if(mark[i])            cout << i << ' ';    cout << '\n';}`