HackerEarth Find the Next! problem solution

In this HackerEarth Find the Next problem solution You are given an array A of length N. For any given integer X, you need to find an integer Z strictly greater than X such that Z is not present in the array A. You need to minimise the value of Z.

HackerEarth Find the Next! problem solution.

#include <bits/stdc++.h>

#define ll long long
using namespace std;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n,q;
    cin>>n>>q;
    int
    ll arr[n+1];
    for(int i=0;i<n;i++) cin>>arr[i];
    map<ll,ll> mp;
    sort(arr,arr+n);
    for(int i=0;i<n;){
        int j=i;
        for(;j<n-1;j++) if(arr[j]<arr[j+1]-1) break;
        for(int k=i;k<=j;k++) mp[arr[k]]=arr[j];
        i=j+1;
    }
    while(q--){
        ll x;
        cin>>x;
        if(mp[x]==0){
            if(mp[x+1]==0) cout<<x+1<<"n";
            else cout<<mp[x+1]+1<<"n";
        } else cout<<mp[x]+1<<"n";
    }
    return 0;
}

Second solution

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

vector<int> Find_It (int Q, vector<int> &A, vector<int> &Queries, int N) {
    sort(A.begin(),A.end());
    map<int,int>ans;
    ans[A[N-1]]=A[N-1]+1;
    for(int i=N-2;i>=0;i--)
    {
        if(binary_search(A.begin(),A.end(),A[i]+1))
            ans[A[i]]=ans[A[i+1]];
        else ans[A[i]]=A[i]+1;
        //cout<<ans[A[i]]<<" ";
    }
    //cout<<"n";
    vector<int>answer;int val;
    for(int i=0;i<Q;i++)
    {
        int x=Queries[i];
        int ind=lower_bound(A.begin(),A.end(),x)-A.begin();
        //cout<<ind<<" ";//int val;
        if(ind==N)
            val=x+1,answer.push_back(x+1);
        else if(A[ind]>x+1)
            val=x+1,answer.push_back(x+1);
        else
            val=ans[A[ind]],answer.push_back(ans[A[ind]]);
        //cout<<val<<"n";
    }
    return answer;
   // Write your code here

}

int main() {

    ios::sync_with_stdio(0);
    cin.tie(0);
    int N;
    cin >> N;
    int Q;
    cin >> Q;
    vector<int> A(N);
    for(int i_A=0; i_A<N; i_A++)
    {
        cin >> A[i_A];
    }
  vector<int> Queries(Q);
    for(int i_Queries=0; i_Queries<Q; i_Queries++)
    {
        cin >> Queries[i_Queries];
    }

    vector<int> out_;
    out_ = Find_It(Q, A, Queries, N);
    for(int i_out_=0; i_out_<out_.size(); i_out_++)
    {
        cout <<out_[i_out_]<<endl;
    }
    return 0;
}

coding problems solutions