【CodeForces 605A】BUPT 2015 newbie practice #2 div2-E

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102419#problem/E

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.

题解：找到这些数中最长的连续的序列（形如123、234、45678），长度为mlen，就是差值为1的上升序列，然后答案就是n-mlen。

因为总共n个数，每个1..n都会出现一次，len[i]表示以i结尾的连续序列的最大长度，则len[a]=len[a-1]+1。

代码：

#include<stdio.h>
int len[100005],n,a,i,mlen=1;
int main(){
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%d",&a);
        len[a]=len[a-1]+1;
        if(len[a]>mlen)mlen=len[a];
    }
    printf("%d
",n-mlen);
    return 0;
}