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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main{
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
	String sequence = scan.nextLine();
	int seqLength = sequence.length();
	int abcVariety = 26;
	int[] eachReqTime =new int[abcVariety];
	int[] eachLastPosition = new int[abcVariety];
	for(int i = 0; i < abcVariety; i++){
	    eachReqTime[i] = 0;
	    eachLastPosition[i] = -1;
	}

	char currentChar;
	int currentIndex;
	for(int i = 0; i < seqLength; i++){
	    currentChar = sequence.charAt(i);
	    currentIndex = (int)currentChar - (int)'a';
	    int currentReqTime = i - 1 - eachLastPosition[currentIndex];
	    eachReqTime[currentIndex] = Math.max(eachReqTime[currentIndex], currentReqTime);
	    eachLastPosition[currentIndex] = i;
	}
	char lastChar = sequence.charAt(seqLength - 1);
	int lastIndex = (int)lastChar - (int)'a';
	for(int i = 0; i < abcVariety; i++){
	    if((i != lastIndex) && (eachLastPosition[i] != -1)){
		int currentReqTime = seqLength - 1 - eachLastPosition[i];
		eachReqTime[i] = Math.max(eachReqTime[i], currentReqTime);
	    }
	}
	int reqTime = seqLength;
	for(int i = 0; i < abcVariety; i++){
	    if(eachLastPosition[i] != -1){
		reqTime = Math.min(reqTime, eachReqTime[i]);
	    }
	}
	System.out.print(reqTime);
    }
}