Problem:
Score : 500 points

Problem Statement

You are given a sequence (P_1,P_2,...,P_N) which is a permutation of the integers from 1 through N.
You would like to sort this sequence in ascending order by repeating the following operation:

Choose an element in the sequence and move it to the beginning or the end of the sequence.

Find the minimum number of operations required.
It can be proved that it is actually possible to sort the sequence using this operation.

Constraints

1 \leq N \leq 2\times 10^5

(P_1,P_2,...,P_N) is a permutation of (1,2,...,N).

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
P_1
:
P_N

Output

Print the minimum number of operations required.

Sample Input 1

4
1
3
2
4

Sample Output 1

2

For example, the sequence can be sorted in ascending order as follows:

Move 2 to the beginning. The sequence is now (2,1,3,4).

Move 1 to the beginning. The sequence is now (1,2,3,4).

Sample Input 2

6
3
2
5
1
4
6

Sample Output 2

4

Sample Input 3

8
6
3
1
2
7
4
8
5

Sample Output 3

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define rep(i,n) for(long long int (i)=0;(i)<(int)(n);(i)++)
#define rrep(i,a,b) for(long long int i=(a);i<(b);i++)
#define rrrep(i,a,b) for(long long int i=(a);i>=(b);i--)
#define all(v) (v).begin(), (v).end()
#define pb(q) push_back(q)
#define Abs(a,b) max(a,b)-min(a,b)
#define YES(condition) if(condition){cout << "YES" << endl;}else{cout << "NO" << endl;}
#define Yes(condition) if(condition){cout << "Yes" << endl;}else{cout << "No" << endl;}
#define Cout(x) cout<<(x)<<endl
#define POSSIBLE(condition) if(condition){cout << "POSSIBLE" << endl;}else{cout << "IMPOSSIBLE" << endl;}
#define Possible(condition) if(condition){cout << "Possible" << endl;}else{cout << "Impossible" << endl;}
#define possible(condition) if(condition){cout << "possible" << endl;}else{cout << "impossible" << endl;}
#define Size(n) (n).size()

typedef long long ll;


using namespace std;

const int INF = 1e9,MOD = 1e9 + 7,ohara=1e6;
const ll LINF = 1e18;

/*---------------------------------------------------------------
long long int kaizyo(long long int hh){
    cmp=1;
    while(hh>1){
        cmp=(cmp*hh)%MOD;
        hh--;
    }
    return cmp;
}
long long int ruizyo(long long int aa, long long int bb){
    if(aa==0){
        return 1;
    }
    else if(aa%2==0){
        long long int tt=ruizyo(aa/2,bb);
        return (tt*tt)%MOD;
    }
    else{
        return (ruizyo(aa-1,bb)*bb)%MOD;
    }
}フェルマ－のア
---------------------------------------------------------------

while(x!=0){
            sum+=x%10;
          / x/=10;
        }
        各桁の和
---------------------------------------------------------------

pair<int,int> p[100000];
cin >> tmp;
p[i]=make_pair(tmp,i);
cout << p[i].second+1 << endl;//ペアの右側つまりiを出力

---------------------------------------------------------------

bool f[100001];//1000000以下の素数を調べよう！
rrep(i,2,100001){
       f[i]=false;
    }
rrep(i,2,100001){
        if(!f[i]){
            for(int j=i+i;j<=100000;j+=i){
                f[j]=true;
            }
        }
    }
    for(int i=3;i<=100000;i+=2){
        if(!f[i]){
            c[i]++;
        }
    }

---------------------------------------------------------------

long long gcd(long long aaa,long long bbb){
    if(bbb==0){
        return aaa;
    }
    return gcd(bbb,aaa%bbb);
}
 
long long lcm(long long aaa,long long bbb){
    long long g = gcd(aaa,bbb);
    return aaa/g * bbb;
}左から最大公約数と最小公倍数


---------------------------------------------------------------

long long int prime_cnt[10000];
       for(int i=2;i*i<=n;i++){
           while(n%i==0){
               n/=i;
               prime_cnt[i]+=1;
           }
           if(n>1){
               prime_cnt[n]+=1;
               break;
           }
      }ある数nを素因数分解しましょう

---------------------------------------------------------------
debug:
cout<<"i: "<<i<<" j: "<<j<<" cnt: "<<cnt<<"\n";

---------------------------------------------------------------*/

long long int n,cnt=0,ans=0,a,b,c,d,cmp,cmpp,m,h,w,x,y,sum[ohara],pos;
int dy[]={1,0,-1,0};
int dx[]={0,1,0,-1};
string alph("abcdefghijklmnopqrstuvwxyz"),s;
bool fl=true;
struct edge{int to,cost;};

//-------------------------↓↓↓↓↓↓------------------------

int main(void){
       cin.tie(0);
    ios::sync_with_stdio(false);

      
      cin>>n;
      pair<ll,ll> p[ohara];
      rep(i,n){
          cin>>a;
          p[i]=make_pair(a,i);
      }
      sort(p,p+n);
      rep(i,n){
          if(p[i].second<p[i+1].second)cnt++;
          else{
              ans=max(ans,cnt);
              cnt=0;
          }
      }
      Cout(n-ans-1);
      

       
return 0;
}