WAP to Find Largest Palindrome In String Difficult One & We Have to Solve it Efficiently

Algorithm

As we know a string can have even & odd length palindrome & any one of them can be larger then other so 1st we will find all odd & even length palindrome of string then we find larger of them.to find odd & even length we will do follow.

for odd length palindrome for each position in string take two variable to store index of element before current index & element after the current index & keep decrementing 1st index & incrementing 2nd index until we won't found any mismatch.save length of maximum odd length palindrome & index of this palindrome. so that in last we just need to compare the maximum odd length palindrome & maximum even lengths palindromic & in last print maximum of them by saving indexes.

class FindAllPalindromes 
{
 public static void main(String[] args)
 {
  FindAllPalindromes finder = new FindAllPalindromes();
  finder.printAllPalindromes("rorradarpottopradarradar");
 }
  
 public void printAllPalindromes(String inputText)
 {
  int cur1=0,max1=-1,cur2=0,max2=-1,a=0,b=0,c=0,d=0;
  if(inputText==null)
  {
   System.out.println("Input cannot be null!");
   return;
  }
  if(inputText.length()<=2)
  {
   System.out.println("Minimum three characters should be present");
  }
  //ODD Occuring Palindromes
  int len = inputText.length();
  for(int i=1;i<len-1;i++){
   for(int j=i-1,k=i+1;j>=0&&k<len;j--,k++){
    if(inputText.charAt(j) == inputText.charAt(k))
    { 
      cur1=k+1-j;
      if(max1<cur1)
      {
        max1=cur1; a=j;b=k+1;
       //System.out.println(inputText.subSequence(j,k+1));
      }
    }else{
     break;
    }
   }
  }
  //EVEN Occuring Palindromes
  for(int i=1;i<len-1;i++)
  {
   for(int j=i,k=i+1;j>=0&&k<len;j--,k++)
   {
    if(inputText.charAt(j) == inputText.charAt(k))
    {
       cur2=k+1-j;
      if(max2<cur2)
      {
        max2=cur2;c=j;d=k+1;
       //System.out.println(inputText.subSequence(j,k+1));
      }
    }
     else
     {
         break;
     }
   }
  }
    if(max1>max2)System.out.println(inputText.subSequence(a,b));
    else System.out.println(inputText.subSequence(c,d));
 }
}

Time Complexity O(N*M)
Space Complexity O(1)
Run Here https://ideone.com/6Zwes

One More Interesting Explanation I Found Here
http://stevekrenzel.com/articles/longest-palnidrome