Problem:
Score : 300 points

Problem Statement

Snuke can change a string t of length N into a string t' of length N - 1 under the following rule:

For each i (1 ≤ i ≤ N - 1), the i-th character of t' must be either the i-th or (i + 1)-th character of t.

There is a string s consisting of lowercase English letters.
Snuke's objective is to apply the above operation to s repeatedly so that all the characters in s are the same.
Find the minimum necessary number of operations.

Constraints

1 ≤ |s| ≤ 100

s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

Print the minimum necessary number of operations to achieve the objective.

Sample Input 1

serval

Sample Output 1

3

One solution is: serval → srvvl → svvv → vvv.

Sample Input 2

jackal

Sample Output 2

2

One solution is: jackal → aacaa → aaaa.

Sample Input 3

zzz

Sample Output 3

0

All the characters in s are the same from the beginning.

Sample Input 4

whbrjpjyhsrywlqjxdbrbaomnw

Sample Output 4

8

In 8 operations, he can change s to rrrrrrrrrrrrrrrrrr.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
 
using namespace std;

char alphas[26] = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

 
int main(){
    string s;
    cin >> s;
    int ans = 1000;
    int siz = s.size();
    for(int i=0;i<27;i++){
        int now = 0;
        int ansnow = 0;
        for(int j=0;j<s.size();j++){
            if(s[j]==alphas[i]){
                if(now!=0){
                    if(s[j]==s[j-1]){
                        now++;
                    }else{
                        ansnow = max(ansnow,j-now-1);
                        now = j;
                    }
                }else{
                    ansnow = max(ansnow,j-now);
                    now = j+1;
                }
            }
        }
        ansnow = max(ansnow,siz-now);
        ans = min(ans,ansnow);
    }

    if(s.size()==1){
        cout << 0 << endl;
    }else{
        cout << ans << endl;
    }
}