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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
 
int main(void){
	char str[100];
	scanf("%s",str);
	int b_letter[26];
	int dif[26];
	int i = 0;
	for(i = 0;i < 26;i++){
		b_letter[i] = 0;
		dif[i] = -1;
	}
	for(i = 0;i < strlen(str);i++){
		int temp = (int)(str[i] - 'a');
		if( i - b_letter[temp] - 1> dif[temp]){
			dif[temp] = i - b_letter[temp] -1;
		}
		b_letter[temp] = i;
		//printf("%d\n",temp);
	}
	int min = 100;
	for(i = 0;i < 26;i++){
		if(strlen(str) - b_letter[i] -1> dif[i])dif[i] = strlen(str) - b_letter[i]-1;
		printf("%d \n",(dif[i]));
		if(dif[i] != -1 && min > dif[i])min = dif[i];
	}
	if(min > (strlen(str) + 1)/2){min = (strlen(str) + 1)/2;}
	printf("%d",min);
	return 0;
}