codeforces 701 c 尺取法

最新推荐文章于 2021-08-02 18:46:52 发布

AC_Lee

最新推荐文章于 2021-08-02 18:46:52 发布

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在一条连续的房屋中，每栋房子里住着不同类型的宝可梦。目标是最少访问多少栋房子以收集所有类型的宝可梦。通过一次不重复访问的方式，找到最小的房子数量。

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题目：

Sergei B., the young coach of Pokemons, has found the big house which consists of n flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number n is only connected with the flat number n - 1.

There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.

Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.

Input

The first line contains the integer n (1 ≤ n ≤ 100 000) — the number of flats in the house.

The second line contains the row s with the length n, it consists of uppercase and lowercase letters of English alphabet, the i-th letter equals the type of Pokemon, which is in the flat number i.

Output

Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.

Examples

input
3
AaA

output
2

input
7
bcAAcbc

output
3

input
6
aaBCCe

output
5

Note

In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.

In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.

In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6.

代码：

#include<set>
#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int n,ans;
set<int> s;
map<int,int> m;
string a;
/*主要是得心细 仔细模拟
尺取法先移动右端点到一个符合条件的情况 然后尝试向前移动左端点
如此反复直到右端点移动到头
*/
int main(){
    ios_base::sync_with_stdio(false);
    cin>>n>>a;
    for(int i=0;a[i]!='\0';i++) s.insert(a[i]-'A');
    int num=s.size();
    ans=n;
    int L=0,R=0,cnt=0;
    while(R<n){
        while(R<n&&cnt<num){///只要区间尚未覆盖所有元素 右端点右移
            if(!m[a[R]]) cnt++;///覆盖新元素
            m[a[R++]]++;///覆盖新元素 右端点右移
        }
        while(m[a[L]]>1) m[a[L++]]--;///尝试向前移动左端点 同时改变区间覆盖情况
        if(cnt==num){///第一个循环有可能出现R=n但还没覆盖满就退出了的情况
            ans=min(ans,R-L);
            //cout<<L<<"  "<<R<<endl;
            m[a[L++]]--;
            cnt--;
        }
    }
    printf("%d\n",ans);
    return 0;
}