cf 888 C K-Dominant Character

最新推荐文章于 2023-10-11 16:26:00 发布

原创最新推荐文章于 2023-10-11 16:26:00 发布 · 251 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#moni

本文介绍了一种算法，用于寻找字符串中最小的k值，使得存在至少一个k-主导字符，即每个长度至少为k的子串都包含该字符。通过计算每个字符的最大间距来确定最小的k值。

C. K-Dominant Character

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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
inputCopy
abacaba
output
2
inputCopy
zzzzz
output
1
inputCopy
abcde
output
3

让求一个最小的切割距离使得每段都有重复的字母，转化为求每个字母最大间距的最小值，注意边界。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <set>
#include <map>
#include <string>

using namespace std;

typedef long long ll;



int main(){
    string s;
    cin>>s;
    int len=s.length();
    map<char,int> mm;
    map<char,int> mmx;

    for(int i=0;i<26;i++){
        char c='a'+i;
        mm[c]=-1;
        mmx[c]=0;
    }

    for(int i=0;i<len;i++){
        mmx[s[i]]=max(mmx[s[i]],i-mm[s[i]]);
        mm[s[i]]=i;
    }
    int ans=len;
    for(int i=0;i<26;i++){
        char c='a'+i;
        //cout<<mmx[c]<<endl;
        mmx[c]=max(mmx[c],len-mm[c]);
        ans=min(ans,mmx[c]);
    }

    cout<<ans<<endl;





    return 0;
}