Educational Codeforces Round 32

最新推荐文章于 2025-07-01 20:10:41 发布

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最新推荐文章于 2025-07-01 20:10:41 发布

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本文介绍了一种算法，用于找出字符串中最小长度k，使得至少存在一个k-主导字符，即每个长度至少为k的子串都包含该字符。通过二分查找优化效率。

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You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least kcontains this character c.

You have to find minimum k such that there exists at least one k-dominant character.

Input

The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 100000).

Output

Print one number — the minimum value of k such that there exists at least one k-dominant character.

Examples

input
abacaba

output
2

input
zzzzz

output
1

input
abcde

output
3

很好的二分宛如智障的我唉

#include<bits/stdc++.h>
using namespace std;
#define maxn 100000+10
char a[maxn];
int len;
int check(int x){
   int c[26+10];
   memset(c,0,sizeof(c));
   for(int j=0;j<x;j++){
      c[a[j]-'a']++;
   }
   int sum=0;
   for(int j=0;j<26;j++){
      if(c[j]) sum++;
      else c[j]=-1;
   }
   for(int j=x;j<len;j++){
      if(c[a[j]-'a']>0){
         c[a[j]-'a']++;
      }
      c[a[j-x]-'a']--;
      if(c[a[j-x]-'a']==0){
        sum--;
      }
      if(sum==0) return 0;
   }
   return 1;
}
int main(){
    cin>>a;
    len=strlen(a);
    int l=0,r=len;
    while(l<r){
        int mid=(l+r)/2;
        if(check(mid)){
            r=mid;
        }
        else {
            l=mid+1;
        }
    }

    cout<<l<<endl;
    return 0;
}