由于必须用两根相等的红色和一根蓝色，因为蓝色+红色一定大于红色，我们只需要考虑两根红色的长度是否大于蓝色（必须满足三角形不等式）;那么就是2*a[l]>b[r] 如果不满足，l++ , 否则配对

#include<bits/stdc++.h>
using namespace std;
#define int long long
int n;
int a[200100],b[200100];
signed main(){
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=1;i<=n;i++)cin>>b[i];
    sort(a+1,a+n+1);
    sort(b+1,b+n+1);
    int ans=0;
    int l=1,r=1;
    while(l<=n&&r<=n){
    if(2*a[l]>b[r]){
        //cout<<"debug"<<a[l]<<" "<<b[r]<<endl;
        ans++;
        l++;
        r++;
    }
    else l++;
    }
    cout<<ans;
}

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll T=1;const ll N=2e5+5;
ll n,x,y;
ll pre[N];ll a[N];ll ans=0;ll k;ll b[N];
void solve(){
	cin>>n;for(ll i=1;i<=n;++i){
		cin>>a[i];a[i]=a[i]*2;
	}
	for(ll i=1;i<=n;++i)cin>>b[i];
	sort(a+1,a+1+n);sort(b+1,b+1+n);
	ll l=1;ll r=1;
	while(l<=n&&r<=n){
		if(a[l]-b[r]>0){
			ans+=1;r++;l++;
		}else{
			l++;
		}
	}cout<<ans;
	
}

int main(){
	
	while(T--)solve();
	return 0;
}