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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 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.
题意:求一个最短区间可以包括所有字符。求最短区间长度。
题解:用尺取法,先取左端点L = 0,然后依次往后取,取到满足题意的结果后,再把左端点往右移,遇到不符合条件的,继续移动右端点,以此类推。在左端点右移的过程中,不断地更新结果ans的最小值即可。
对于大小写的问题就不用分开了,一个数组记录下全部的大写小写。
代码如下:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MAX 100000
#define idx(x) (x-'A')
char c[MAX+11];
int vis[58]; //0表示A,57表示z
int main()
{
int n;
while (~scanf ("%d",&n))
{
scanf ("%s",c);
int dif = 0;
CLR(vis,0);
for (int i = 0 ; i < n ; i++)
if (!vis[c[i]-'A'])
{
dif++;
vis[c[i]-'A']++;
}
int i,j;
int ans = INF;
int ant = 0;
CLR(vis,0);
int st = 0;
for (i = 0 ; i < n ; i++)
{
int id = idx(c[i]);
if (!vis[id])
ant++;
vis[id]++;
if (ant == dif) //首部前移
{
while (ant == dif)
{
ans = min (ans , i - st + 1);
int t = idx(c[st]);
if (vis[t] == 1) //此次前移就退出了
ant--;
vis[t]--;
st++;
}
}
}
printf ("%d\n",ans);
}
return 0;
}