Java Program to Find All Palindromic Sub-Strings of a String

In this problem, we are given a String and our task is to find all palindromic sub-strings of the specified length. There are two ways to solve the problem. The first way is to compare the characters of sub-string from start to last, and another way is to reverse the sub-string and compare it with the original sub-string to check whether it is palindrome.

String in Java is a class which represents a sequence of characters. It is immutable which means once it is created a String object cannot be changed. And, sub-string is a small portion or part of a string.

Example Scenario 1:

Input: string str = "abcd"
Output: res = 4

The palindromic substrings are 'a', 'b', 'c', and 'd'.

Example Scenario 2:

Input: string str = "abbab"
Output: res = 8

The palindromic substrings are 'a', 'abba', 'b', 'bb', 'b', 'bab', 'a', 'b'.

Using for Loop

In this approach, we will use the for loop to get all sub-strings of the given string. We will match the characters of the sub-string from the start and end to check whether the string is a palindrome. If any character doesn't match, we can say the string is not palindromic and move ahead.

Example

The following Java program shows how to find palindromic sub-strings of a given string using for loop.

import java.io.*;

public class Main {
   public static boolean isPalindrome(String temp) {
      int len = temp.length();
      // Match string characters from the start and end
      for (int m = 0; m < len / 2; m++) {
         if (temp.charAt(m) != temp.charAt(len - m - 1)) {
            return false;
         }
      }
      return true;
   }
   public static void main(String[] args) {
      // Custom input string
      String alpha = "abcd";
      // To store the total number of palindromic substrings
      int palCounts = 0;
      // Get all substrings starting with index p.
      for (int p = 0; p < alpha.length(); p++) {
         // Get all substrings ending at q index
         for (int q = p; q < alpha.length(); q++) {
            // Get Substring
            String subStr = alpha.substring(p, q + 1);
            // If the string is a palindrome, increment count
            if (isPalindrome(subStr)) {
               palCounts++;
            }
         }
      }
      System.out.println("The total number of palindromic substrings of the given string is " + palCounts);
   }
}

On running this code, it will generate the following result

The total number of palindromic substrings of the given string is 4

Following are the time and space complexities

//To find all substrings and checking whether it is palindromic.
Time complexity ? O(N^3)
//To store the substring
Space complexity ? O(N)

Using while Loop

In this approach, we will use the while loop to get all sub-strings of the given string. After that, we will use the reverse() method to reverse the string and compare it with the original sub-string to check if it is palindromic or not.

Example

In this Java program, we use the while loop to print find palindromic sub-strings of a given string.

import java.io.*;

public class Main {
   public static void main(String[] args) {
      // Custom input string
      String alpha = "abcd";
      // To store the total number of palindromic substrings
      int palCounts = 0;
      int p = 0;
      // Get all substrings starting with index p.
      while (p < alpha.length()) {
         int q = p;
         // Get all substrings ending at q index
         while (q < alpha.length()) {
            // Get Substring
            String subStr = alpha.substring(p, q + 1);
            // Reverse substring
            String rev = new StringBuffer(subStr).reverse().toString();
            // If substring and its reverse are equal
            if (subStr.equals(rev)) {
               palCounts++;
            }
            q++;
         }
         p++;
      }
      System.out.println("The total number of palindromic substrings of the given string is " + palCounts);
   }
}

Output of the above code is as follows ?

The total number of palindromic substrings of the given string is 4

Following are the time and space complexities

//To get all substrings and reverse each substring
Time complexity ? O(N^3)
//To store the reversed substring
Space complexity ? O(N)