Codeforces Round #364 (Div. 2) C. They Are Everywhere （窗口滑动）

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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 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 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.

题目大意：

给定一个长为n的字符串，然后找出最短的一个字串包含母串中所有种类的字母。

题目解析：

字符串长度为100000，n^2肯定是不可能的，所以要考虑其他做法，窗口滑动是使用一个双指针来实现相应字符串的匹配

起始时L和R都在0，然后随着匹配R一直向后移动，直到符合条件为止，更新一下答案，然后开始移动L，ans-=1，直到不符合条件。

再次开始R的移动，直到整个字符串都查找完，第三组样例很好！

#include <algorithm>
#include <iostream>
#include <numeric>
#include <cstring>
#include <iomanip>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const LL Max = 405;
const double esp = 1e-6;
const double PI = 3.1415926535898;
const int INF = 0x3f3f3f3f;
using namespace std;

char arr[100005];
map<char,int>M;

int main(){
    int n,len;
    while(~scanf("%d %s",&n,arr)){
        len = 0;
        for(int i=0;i<n;i++){
            if(M[arr[i]] == 0)
                len += 1;
            M[arr[i]] += 1;
        }
        M.clear();

        int l = 0,r = 0,num = 0,ans = n;

        while(r <= n-1){
            if(num < len){
                if(M[arr[r]] == 0)
                    num += 1;
                M[arr[r]] += 1;r += 1;
            }
            else{
                ans = min(ans,r-l);
                if(M[arr[l]] == 1)
                    num -= 1;
                M[arr[l]] -= 1;l += 1;
            }
        }
        while(r - l >= len){
            if(num == len)
                ans = min(ans,r-l);
            if(M[arr[l]] == 1)
                num -= 1;
            M[arr[l]] -= 1;l += 1;
        }
        printf("%d\n",ans);
    }
    return 0;
}