Codeforces Round #364 (Div. 2) C. They Are Everywhere(stl+乱搞)

C. They Are Everywhere

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Sergei B., the young coach of Pokemons, has found the big house which consists ofn 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 number1 is only connected with the flat number2 and the flat numbern is only connected with the flat numbern - 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 lengthn, it consists of uppercase and lowercase letters of English alphabet, thei-th letter equals the type of Pokemon, which is in the flat numberi.

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

3AaA

Output

2

Input

7bcAAcbc

Output

3

Input

6aaBCCe

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.

用queue+set+map 队列中存放当前的取到的元素，set存放元素的种类，map存储队列中的元素各出现了多少个。需要预处理出元素一共有多少类(K)。如果队列中的元素种类不足K那么下一个元素入队，如果队中的元素种类满足K个，那么首先要判断是否为最小值，然后需要判断队首元素是否为队列中唯一这一种类的元素，如果唯一，则要在集合删掉这一元素，并且让其出队，若不是，则这一元素在这队中这一种类的数目减一然后出队。

用双指针+二分也可以实现该过程。但是我感觉比较麻烦。

总之时间复杂度为O(nlogn);

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<string>#include<set>#include<map>#include<queue>using namespace std;set<char>q;map<char,int>p;set<char>q2;queue<char>qu;char str[100005];int main(){    int n;    scanf("%d",&n);scanf("%s",str);    q.clear();q2.clear();p.clear();    int minn=100005;    for (int i=0;i<n;i++) q.insert(str[i]);    int kind=q.size();    int flag=1;    for (int j=-1;j<n;)    {       if (q2.size()!=kind) {j++;       if (flag) {q2.insert(str[j]);qu.push(str[j]);p[str[j]]++;}}       if (j==n-1) flag=0;       if (q2.size()==kind) {            char fr=qu.front();           if (qu.size()<minn) minn=qu.size();           if (p[fr]==1) {q2.erase(fr); p[fr]--;qu.pop();}           else {p[fr]--;qu.pop();}        }    }    printf("%d",minn);}

如果满足K个，那么需要