# HackerEarth Missing Number problem solution

In this HackerEarth Missing Number problem solution, You are given an array A. You can decrement any element of the array by 1. This operation can be repeated any number of times. A number is said to be missing if it is the smallest positive number which is a multiple of 2 that is not present in the array A. You have to find the maximum missing number after all possible decrements of the elements.

## HackerEarth Missing Number problem solution.

`#include<bits/stdc++.h>#include <stdint.h>using       namespace       std;#define     ff          first#define     ss          second#define     pb          push_back#define     mp          make_pair#define     eps         1e-9#define     inf         1e18#define     PI          3.14159265#define     mset(A,num,n);    memset(A,num,n);#define     all(a)        a.begin(),a.end()#define     allr(a)       a.rbegin(),a.rend()#define     nl          cout << endl#define     vi          vector<int>#define     vll         vector<ll>#define     pi          pair<int,int>//#define     trace1(x)               cout <<#x<<": "<<x<< endl;#define     trace2(x, y)              cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<< endl;#define     trace3(x, y, z)           cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;#define     trace4(a, b, c, d)        cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;#define     trace5(a, b, c, d, e)     cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<<": "<<e<<endl;const int     MOD =         1e9+7;const int     N   =         1e6+5;typedef     long long int     ll;typedef     int64_t       lln;ll POWER[65];ll aonb(ll a, ll b) {ll ret=1;while(b) {if(b&1) ret*=a;a*=a;if(ret>=MOD) ret%=MOD;if(a>=MOD) a%=MOD;b>>=1;}return ret;}void precompute() {  POWER[0]=1;  for(int i=1;i<63;i++) POWER[i]=POWER[i-1]<<1LL;}int main(){  std::ios_base::sync_with_stdio(false);  int start_s=clock();  int t;  cin >> t;  while(t--){    int n;    cin >> n;    vector<int> A(n);    for (int i=0;i<n;i++)      cin >> A[i];          sort(A.begin(),A.end());    int current=2;    for (int i=0;i<n;i++)    {      if (A[i]>=current)      {        current+=2;      }    }    cout << current << endl;  }     int stop_s=clock();//  cout << "time: " << (stop_s-start_s)/double(CLOCKS_PER_SEC)*1000 << endl;    return 0;}`

### Second solution

`#include <bits/stdc++.h>#define ll long long#define ull unsigned long long#define pb push_back#define mp make_pair#define fi first#define se second#define be begin()#define en end()#define all(x) (x).begin(),(x).end()#define alli(a, n, k) (a+k),(a+n+k)#define REP(i, a, b, k) for(__typeof(a) i = a;i < b;i += k)#define REPI(i, a, b, k) for(__typeof(a) i = a;i > b;i -= k)#define REPITER(it, a) for(__typeof(a.begin()) it = a.begin();it != a.end(); ++it)#define y0 sdkfaslhagaklsldk#define y1 aasdfasdfasdf#define yn askfhwqriuperikldjk#define j1 assdgsdgasghsf#define tm sdfjahlfasfh#define lr asgasgash#define norm asdfasdgasdgsd#define have adsgagshdshfhds#define eps 1e-6#define pi 3.141592653589793using namespace std;template<class T> inline T gcd(T a, T b) { while(b) b ^= a ^= b ^= a %= b; return a; }template<class T> inline T mod(T x) { if(x < 0) return -x; else return x; }typedef vector<int> VII;typedef vector<ll> VLL;typedef pair<int, int> PII;typedef pair<ll, ll> PLL;typedef pair<int, PII > PPII;typedef vector< PII > VPII;typedef vector< PPII > VPPI;const int MOD = 1e9 + 7;const int INF = 1e9;const int MAX = 1e5 + 5;int a[MAX];int main(int argc, char* argv[]){    if(argc == 2 or argc == 3) freopen(argv[1], "r", stdin);    if(argc == 3) freopen(argv[2], "w", stdout);    ios::sync_with_stdio(false);    int t, n, x;    cin >> t;    assert(1 <= t and t <= 100);    while (t--) {        cin >> n;        assert(1 <= n and n <= 100000);        REP(i, 0, n, 1) {            cin >> a[i];            assert(0 <= a[i] and a[i] <= INF);        }        sort(alli(a, n, 0));        x = 2;        REP(i, 0, n, 1) {            if (a[i] >= x) {                x += 2;            }        }        cout << x << endl;    }    return 0;}`