Code: Search a Reverse Substring

 You are given a String input. You are to find the longest substring of input such that the reversal of the substring is also a substring of input. In case of a tie, return the string that occurs earliest in input.

Notes 
    -The substring and its reversal may overlap partially or completely. 
    -The entire original string is itself a valid substring (see example 4).
Constraints 
    -input will contain between 1 and 50 characters, inclusive. 
    -Each character of input will be an uppercase letter ('A'-'Z').
Examples
0) "XBCDEFYWFEDCBZ" 
    Returns: "BCDEF" 
    We see that the reverse of BCDEF is FEDCB, which appears later in the string.
1) "XYZ" 
    Returns: "X" 
    The best we can do is find a one character substring, so we implement the tie-breaker rule of taking the earliest one first.
2) "ABCABA" 
    Returns: "ABA" 
    The string ABA is a palindrome (it's its own reversal), so it meets the criteria.
3) "FDASJKUREKJFDFASIREYUFDHSAJYIREWQ" 
    Returns: "FDF"
    The similar as above.
4) "ABCDCBA" 
    Returns: "ABCDCBA" 
    Here, the entire string is its own reversal.

class  ReverseSubstring
{
    static void Main(string [] args)
     {
        string[] strings = new string[]  { "XBCDEFYWFEDCBZ", "XYZ" , 
                "FDASJKUREKJFDFASIREYUFDHSAJYIREWQ", "ABCDCBA" } ;
        for ( int i=0 ; i < strings.Length ; ++ i )
         {
            System.Console.WriteLine(FindReversed(strings[i]));
        }
    }

     private static string FindReversed(string  input)
     {
        string substring = string .Empty;
        for ( int i=0 ; i < input.Length ; ++ i )
         {
            for ( int j=0 ; j < input.Length ; ++ j )
             {
                if ( input[i] == input[input.Length-1- j] )
                 {
                    if ( i == input.Length-1- j )
                     {
                        if ( substring.Length == 0  )
                         {
                            substring =  input[i].ToString();
                        }
                         break ;
                    }
                     else
                     {
                        int k = 0 ;
                        for ( ; k < input.Length-i-j ; ++ k )
                         {
                            if ( input[i+k] != input[input.Length-1-j-k] ) break ;
                        }
                        if ( substring.Length <  k )
                         {
                            substring =  input.Substring(i, k);
                        }
                    }
                }
            }
        }
         return  substring;
    }
}

BCDEF
X
ABA
FDF
ABCDCBA