Why/how does the Longest Proper Prefix/Suffix algorithm work?

Question

The LPS (Longest Proper Prefix which is also a Suffix) algorithm goes as follows:

public static int[] constructLPSArray(String s) {
        int n = s.length();
        int[] arr = new int[n];
        int j = 0;
        for (int i = 1; i < n; ) {
            if (s.charAt(i) == s.charAt(j)) {
                arr[i] = j + 1;
                i++;
                j++;
            } else {
                if (j != 0) {
                    j = arr[j - 1];
                } else {
                    i++;
                }
            }
        }
        return arr;
    }

The if (s.charAt(i) == s.charAt(j)) part looks clear, but the else part is unclear. Why do we do:

if (j != 0) {
  j = arr[j - 1];
} else {
  i++;
}

More specifically, why does j = arr[j - 1] work ? Or why do we even do it? How do we validate the correctness of this step?

Answer 1

Let's say we are parsing an array of characters with i and j positioned like this:

a b a b x x a b a b ...
      ^           ^
      j           i

with arr holding:

0 0 1 2 0 0 1 2 3 4

i. e., the length of the longest prefix/suffix of each substring of s of that length until i. You can probably guess how that was generated from the rest of the algorithm. Now, if the next character after i does not match the next character after j,

a b a b x x a b a b a ...
        ^           ^
        j           i

we don't have to retry the matching, because we know the longest prefix/suffix of our previous prefix/suffix! Looking up arr[j - 1] yields 2 – so we essentially cached the information that the parts highlighted here

A B a b x x a b A B a ...
=== ^           === ^
    j               i

are identical, and don't need to be compared again!

Answer 2

  *Here's one more solution*
  int length=str.length();
  int mid=length/2;
  if(length<2){
       System.out.println("-1");
  }
  for(int i=mid;i>=0;i--){
      String prefix=str.substring(0,i);
      String suffix=str.substring(length-i,length);
      if(suffix.equals("") || prefix.equals("")){
            System.out.println("-1");
      }
      if(suffix.equals(prefix)){
          System.out.println(suffix.length());
          break;
      }
      
  }