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;
#define REP(i, n) for(int i = 0; i < n; i++)

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    string s;
    cin >> s;
    map<char, int> mp;
    REP(i, s.size()) mp[s[i]]++;

    int ma = 0;
    for(auto x : mp) ma = max(ma, x.second);

    if(ma == s.size()){
        cout << 0 << endl;
        return 0;
    }
    if(ma == 1){
        cout << max((int)(s.size() - 3), 1) << endl;
        return 0;
    }

    vector<char> C;
    for(auto x : mp){
        if(x.second == ma){
            C.push_back(x.first);
        }
    }

    int mi = 100;
    REP(i, C.size()){
        string a = s, b = "";
        while(true){
            int cnt = 0;
            REP(j, a.size() - 1){
                if(a[j] == C[i] or a[j + 1] == C[i]){
                    b += C[i];
                    cnt++;
                }else b += a[j];
            }
            set<char> st;
            REP(j, a.size()) st.insert(a[j]);
            if(st.size() == 1){
                mi = min(mi, (int)(s.size() - a.size()));
                break;
            }
            a = b;
            b = "";
        }
    }
    cout << mi << endl;

    return 0;
}