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 (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

typedef long long ll;
typedef char keytype;

int gcd(int a, int b){ return (a % b) ? gcd(b, a % b) : b;}
int lcm(int a, int b){ return (a / gcd(a, b)) * b;}

/* a[0], ..., a[n-1] の数の最大公約数 */
int ngcd(int n, int a[])
{
    int i, d;

    d = a[0];
    for (i = 1; i < n && d != 1; i++)
        d = gcd(a[i], d);
    return d;
}

void quicksort(keytype a[], int first, int last)
{
    int i, j;
    keytype x, t;

    x = a[(first + last) / 2];
    i = first;  j = last;
    for ( ; ; ) {
        while (a[i] < x) i++;
        while (x < a[j]) j--;
        if (i >= j) break;
        t = a[i];  a[i] = a[j];  a[j] = t;
        i++;  j--;
    }
    if (first  < i - 1) quicksort(a, first , i - 1);
    if (j + 1 < last) quicksort(a, j + 1, last);
}

int is_prime(int x){
    if(x <= 1) return 0;
    for(int i = 2; i * i <= x; i++){
        if(x % i == 0) return 0;
    }
    return 1;
}

int is_substr(char s[], char t[], int i){
    int flag = 1;

    if(s[i] == '\0') flag = 0;

    for(int j = 0; j < strlen(t); ++j, ++i){
        if(s[i] != t[j]){
            flag = 0;
            break;
        }
    }
    return flag;
}

// factor list, number of factors, natural number
void factorize(int ftr[], int *num, int n){
    int a = 2, cnt = 0;

    while(n >= a){
        if(n % a == 0){
            n = n / a;
            ftr[cnt++] = a;
        }else{
            a++;
        }
    }
    *num = cnt;
}

//#define N 5
//int p[N + 1];
int nextperm(int p[], int N)  /* 辞書式順序で次の順列 */
{
    int i, j, t;

    i = N - 1;
    p[0] = 0;  /* 番人 */
    while (p[i] >= p[i + 1]) i--;
    if (i == 0) return 0;  /* 完了 */
    j = N;
    while (p[i] >= p[j]) j--;
    t = p[i];  p[i] = p[j];  p[j] = t;
    i++;  j = N;
    while (i < j) {
        t = p[i];  p[i] = p[j];  p[j] = t;  i++;  j--;
    }
    return 1;  /* 未了 */
}

void comb(int size1, int size2, int v[size1][size2]){
  for(int i = 0; i < size1; i++){
    v[i][0] = 1;
    v[i][i] = 1;
  }
  for(int k = 1; k < size1; k++){
    for(int j = 1; j < k; j++){
      v[k][j] = (v[k-1][j-1] + v[k-1][j]);
    }
  }
}

int abs_val(int a, int b){
    if(a > b) return a - b;
    else return b - a;
}

const int inf = 1012345678;
int main(){
    int n;
    int p[200005];

    scanf("%d", &n);
    for(int i = 0; i < n; ++i){
        scanf("%d", &p[i]);
    }

    int max = 1;
    int cnt = 1;
    for(int i = 0; i < n; ++i){
        cnt = 1;
        for(int j = i + 1; j < n; ++j){
            if(p[i] + cnt == p[j]){
                cnt++;
            }
        }
        if(max < cnt){
            max = cnt;
        }
    }
    int ans = n - max;

    printf("%d\n", ans);

    return 0;
}