How to solve this problem of strings and keys?

An Ethical Hacker has been tasked to find vulnerabilities in Transaction Operations of an
Organization XYZ. He has been able to find a relation between the Transaction ID and the Key generated
by the System. Write a program that generates Key from the Transaction ID.
Transaction ID S is a String consisting of only lower case alphabetic characters, i.e. S[i] ∈ [a, z]. K
is said to be a possible key of S, if there exist at least one-character C ∈ [a, z] in each and every substring
of S of length K. Key of ID is the minimum of such possible Ks.
Input
First line contains Transaction ID S (1≤ length(S)≤ 100000)
Output
Key of Transaction ID i.e. Minimum of possible keys Ks
Like if inputs are:-
zbcde   /*here c is common between substrings 
        Having 3 max length i.e zbc cde
bbbbb    /* in this only b is there so length 
         Becomes 1 i.e b
zbzczbz  /*here z is common in maximum length 
         2 of given string i.e zb bz zc cz zb 
          bz 
Then output will be like:-
3
1
2

<pre>import java.util.Scanner;

public class lab3hw {
	public static boolean aresame(int[] array)
	{
	    boolean fe = array[0] == 0;
	    for(int i = 1; i < array.length; i++)
	    {
	        if(fe)
	            if(array[i] != 0) return false;
	        else 
	            if(!(array[0]==array[i])) return false;
	    }

	    return true;
	}
	public static boolean samechar(String s)
	{
	    int n = s.length();
	    boolean boo=false;
	    for (int i = 1; i < n; i++)
	        if (s.substring(i,i+1).equals(s.substring(0,1))) {
	            boo=false;
	            
	        }else {
	        	boo=true;
	        }
	    return boo;
	}
	public static int conditions(String str) {  
	 int ci, i, j, k, l=0;
	 int result=0;
     char c, ch;
     i=str.length();
     if(samechar(str)==false) {
    	 
    	 result=1;
     }else {
	     for(c='A'; c<='z'; c++)
	     {
	         k=0;
	         for(j=0; j<i; j++)
	         {
	             ch = str.charAt(j);
	             if(ch == c)
	             {
	                 k++;
	             }
	         }
	         if(k>0)
	         {
	             int[]n=new int[50];
	             for(int m=0;m<i;m++){
	             n[m]=k;
	             }
	         if(aresame(n)==true) { 
	        	 if((str.length())%2==0) {
	        		 result= str.length()/2;}
	        	 else {
	        		 result=(str.length()+1)/2;
	        	 }
	         }else if(aresame(n)==false) {
	        	 
	         }
	         }
	     }
	     
     }
     return result;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner a=new Scanner(System.in);
		String b=a.next();
		System.out.println(conditions(b));
		
	}

}