350A - TL - CodeForces Solution

brute force greedy implementation *1200

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Python Code: 

n, m = map(int,input().split(" "))
correct = list(map(int, input().split(" ")))
wrong = list(map(int, input().split(" ")))

min_possible_v = 2*(min(correct))
min_correct = max(correct)
max_possible = min(wrong)

if min_correct >= max_possible:
    print(-1)
else:
    if min_possible_v >= max_possible:
        print(-1)
    else:
        print(max(min_possible_v, min_correct))

C++ Code: 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define N 100001
        ll prime[N];
        vector<int>all_primes;
        ll prime_factors[N];
        void seive()
        {
            memset(prime,0,sizeof(prime));
            prime[0]=1;prime[1]=1;
            for(ll i=2;i*i<=N-1;i++)
            {
            if(prime[i]==0)
                {
                    for(ll j=i*i;j<N;j+=i)
                    {
                        prime[j]=1;
                    }
                }
            }
            for(int i=2;i<N;i++)
            {
                if(prime[i]==0)
                    all_primes.push_back(i);
            }
        }

        void spf()
        {
            for(ll i=1;i<N;i++)
            {
                prime_factors[i]=i;
            }

            for(ll i=2;i*i<=N-1;i++)
            {
                for(ll j=i*i;j<=N-1;j+=i)
                {
                    if(prime_factors[j]==j)
                        prime_factors[j]=i;
                }
            }
        }

int main()
{
    int n,m;
    cin>>n>>m;
    int a[n],b[m];
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    for(int i=0;i<m;i++)
    {
        cin>>b[i];
    }
    sort(a,a+n);
    sort(b,b+m);
    int ans=-1;
    if(n==1 && (a[0]*2)<b[0])
        ans=a[0]*2;
    else if(a[n-1]<b[0] && a[n-1]>=(2*a[0]))
        ans=a[n-1];
    else if((a[0]*2)>=a[n-1] && (a[0]*2)<b[0])
        ans=a[0]*2;

    cout<<ans;
}


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