如果將1和3都放到正確的位置，2自然也在正確的位置。那麼統計1,2,3的數量num1,num2,num3。再看前num1個數有幾個（設x個）不是1，那麼x個1肯定要移。設前num1個數有y個3，最後num3個數有z個1，那麼這個過程中，最多能將min(y,z)個3移到正確位置。剩下num3-min(y,z)個3也必須移到最後num3個位置

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include 
 
<iomanip>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <cassert>
#include <stack>
#define mkp make_pair
using namespace std;
const double EPS=1e-8;
typedef long long lon;
const lon SZ=1060,INF=0x7FFFFFFF;
int arr[SZ];

int main()
{
    std::ios::sync_with_stdio(
 
0);
    //freopen("d:\\1.txt","r",stdin);
    lon casenum;
    //cin>>casenum;
    //for(lon time=1;time<=casenum;++time)
    {
        int n;
        cin>>n;
        for(int i=1;i<=n;++i)cin>>arr[i];
        int num1=count(arr+1,arr+1+n,1);
        int num2=count(arr+1,arr+1+n,2);
        
 
int num3=count(arr+1,arr+1+n,3);
        int x=0,y=0,z=0,w=0;
        for(int i=1;i<=num1;++i)
        {
            if(arr[i]!=1)++x;
            if(arr[i]==3)++z;
        }
        for(int i=num1+num2+1;i<=n;++i)
        {
            if(arr[i]!=3)++y;
            if(arr[i]==1)++w;
        }
        //cout<<num1<<" "<<num2<<" "<<num3<<endl;
        //cout<<x<<" "<<y<<" "<<z<<endl;
        cout<<(x+y-min(z,w))<<endl;
    }
    return 0;
}