Description

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.

题意:就是寻找最小的区间,它能包含该字符串所有字母

我们设立起点和终点,l,r;

l从该字母第一次出现开始,如果后面该字母再次出现,就移动到另外一个字母第一次出现的位置

r从头遍历到尾,如果遍历符合条件就更新一次

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<vector>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
using namespace std;
int flag[100005];
int main()
{
set<char>q;
int a[100005];
string s;
int n;
int sum;
int l=0,r=0;
int len=(1<<30);
cin>>n;
cin>>s;
for(int i=0;i<n;i++)
{
q.insert(s[i]);
}
sum=q.size();
for(int r=0;r<n;)
{
int cot=s[r]-'0';
flag[cot]++;
if(flag[cot]==1)
{
sum--;
}
// cout<<flag[cot]<<"B"<<endl;
//cout<<flag[s[l]
while(flag[s[l]-'0']>1)
{
flag[s[l]-'0']--;
l++;
// cout<<l<<"A"<<endl;
}
if(sum==0)
{
len=min(len,r-l+1);
// cout<<r<<endl;
// cout<<l<<endl;
}
r++;
// cout<<r<<"A"<<endl;
}
cout<<len<<endl;
return 0;
}

  

Codeforces Round #364 (Div. 2) C的更多相关文章

  1. Codeforces Round #364 (Div. 2)

    这场是午夜场,发现学长们都睡了,改主意不打了,第二天起来打的virtual contest. A题 http://codeforces.com/problemset/problem/701/A 巨水无 ...

  2. Codeforces Round #364 (Div.2) D:As Fast As Possible(模拟+推公式)

    题目链接:http://codeforces.com/contest/701/problem/D 题意: 给出n个学生和能载k个学生的车,速度分别为v1,v2,需要走一段旅程长为l,每个学生只能搭一次 ...

  3. Codeforces Round #364 (Div.2) C:They Are Everywhere(双指针/尺取法)

    题目链接: http://codeforces.com/contest/701/problem/C 题意: 给出一个长度为n的字符串,要我们找出最小的子字符串包含所有的不同字符. 分析: 1.尺取法, ...

  4. 树形dp Codeforces Round #364 (Div. 1)B

    http://codeforces.com/problemset/problem/700/B 题目大意:给你一棵树,给你k个树上的点对.找到k/2个点对,使它在树上的距离最远.问,最大距离是多少? 思 ...

  5. Codeforces Round #364 (Div. 2) B. Cells Not Under Attack

    B. Cells Not Under Attack time limit per test 2 seconds memory limit per test 256 megabytes input st ...

  6. Codeforces Round #364 (Div. 2) Cells Not Under Attack

    Cells Not Under Attack 题意: 给出n*n的地图,有给你m个坐标,是棋子,一个棋子可以把一行一列都攻击到,在根据下面的图,就可以看出让你求阴影(即没有被攻击)的方块个数 题解: ...

  7. Codeforces Round #364 (Div. 2) Cards

    Cards 题意: 给你n个牌,n是偶数,要你把这些牌分给n/2个人,并且让每个人的牌加起来相等. 题解: 这题我做的时候,最先想到的是模拟,之后码了一会,发现有些麻烦,就想别的方法.之后发现只要把它 ...

  8. Codeforces Round #364 (Div. 2)->A. Cards

    A. Cards time limit per test 1 second memory limit per test 256 megabytes input standard input outpu ...

  9. Codeforces Round #364 (Div. 2) E. Connecting Universities

    E. Connecting Universities time limit per test 3 seconds memory limit per test 256 megabytes input s ...

  10. Codeforces Round #364 (Div. 2) C.They Are Everywhere

    C. They Are Everywhere time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

随机推荐

  1. tarjan求割点

    首先给大家一个网址讲的比较细:http://www.cnblogs.com/en-heng/p/4002658.html 如果还有不懂的话,可以回来再看看我的文章; 概念明确: 树边:(在[2]中称为 ...

  2. URI is not registered (Settings | Languages & Frameworks | Schemas and DTDs)

    解决:鼠标悬于上方Alt + Enter 选择Ignore

  3. Ubuntu 14.04开发环境

    安装ssh服务:sudo apt-get install openssh-server 安装vim:sudo apt-get install vim-gtk 安装gparted:sudo apt-ge ...

  4. 使用Visual Studio进行单元测试-Part5

    本文主要介绍Visual Studio(2012+)单元测试框架的一些技巧: 如何模拟类的静态构造函数 如何测试某方法被调用过 如何测试某方法执行的次数 并行编程测试注意事项 一.如何模拟类的静态构造 ...

  5. node.js的国内源

    node.js在使用npm安装包是,由于源是国外的,有可能会被GFW屏蔽. 通过下面的方法可以把源指向国内的. 具体方法如下: 编辑 ~/.npmrc 加入下面内容 registry = http:/ ...

  6. 删除老的Azure Blob Snapshot

    客户有这样的需求:每天需要对VM的数据进行备份,但如果备份的时间超过一定的天数,需要进行清除. 本文也是在前一篇Azure Blob Snapshot上的优化. "Azure blob St ...

  7. js基础之变量类型

    1.NAN(Not a number) 不是一个数字 自身:console.log(NaN==NaN)和console.log(NaN===NaN)返回值都是false; 其他函数,isNaN()可用 ...

  8. javaScript之this的五种情况

    this一直是JavaScript研究的难题,特别是在笔试和面试中的各种程序分析问题中,也常常会被问到.下面来看一看this被运用的五中情况: (1)       纯粹的函数调用 函数最普通用法,此时 ...

  9. Sequence Models 笔记(二)

    2 Natural Language Processing & Word Embeddings 2.1 Word Representation(单词表达) vocabulary,每个单词可以使 ...

  10. TCP/IP四层体系结构

    1.数据链路层  2.网络层  3.传输层  4.应用层 , 其中IP是在第二层网络层中,TCP是在第3层传输层中, Internet体系结构最重要的是TCP/IP协议,是实现互联网络连接性和互操作性 ...