CodeForces 606C--Sorting Railway Cars，思路题~~~

C - Sorting Railway Cars

  Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u    

CodeForces 606C

Description

An infinitely long railway has a train consisting of n cars, numbered from 1 to n (the numbers of all the cars are distinct) and positioned in arbitrary order. David Blaine wants to sort the railway cars in the order of increasing numbers. In one move he can make one of the cars disappear from its place and teleport it either to the beginning of the train, or to the end of the train, at his desire. What is the minimum number of actions David Blaine needs to perform in order to sort the train?

Input

The first line of the input contains integer n (1 ≤ n ≤ 100 000) — the number of cars in the train.

The second line contains n integers p_i (1 ≤ p_i ≤ n, p_i ≠ p_j if i ≠ j) — the sequence of the numbers of the cars in the train.

Output

Print a single integer — the minimum number of actions needed to sort the railway cars.

Sample Input

Input

5
4 1 2 5 3

Output

2

Input

4
4 1 3 2

Output

2

Hint

In the first sample you need first to teleport the 4-th car, and then the 5-th car to the end of the train.

     做过好几遍的题了，题目还不错，一前以为用单调递增子序列做，后来发现求出连续递增的最长长度即可，就像1 2 3  4 5这种， 仔细看看题多看几遍就可以发现题意，然后转化：

    用另外一个数组储存最长序列长度，只要在输入得时候判断比它小一的元素是否在其左边，是，则其下标为左边小一的元素下标加一，否，则为一，其实思路清晰甚至都不用判断，直接输入a[i],然后下标b[a[i]]=b[a[i]-1]+1；

     

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=100000+10;
int a[N],b[N];
int main()
{
    int n,i;
    while(~scanf("%d",&n))
    {
        memset(b,0,sizeof(b));
        for(i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            b[a[i]]=b[a[i]-1]+1;
        }
        sort(b+1,b+n+1);
        if(b[n]==n)
            printf("0\n");
        else
            printf("%d\n",n-b[n]);
    }
    return 0;
}