Write program to palindrome string

//Method 1

 public void PalindromeString{

//main method

public static void main(String args[])

{

//String declare and initiation 

 String s="MADAM";

String rev="";

for(int i=s,lenght()-1;i>=0;i--){

  rev=rev+s.charAt(i);

}

      if(s.equals(rev)){

       System.out.println("this is palindrome string");//MADAM

       }

      else{

        System.out.println("This is not palindrome screen"); //!=MADAM

       }

  }

}

=====================================================

🔹 Step 1: Start with the Concept

• A palindrome string is a string that reads the same forward and backward (e.g., "MADAM", "LEVEL").

• The logic is to reverse the string and compare it with the original.

🔹 Step 2: Walk Through the Code

• String Declaration:

  java

  String s = "MADAM";
  

  → This is the input string we want to check.

• Reverse Logic:

  java

  String rev = "";
  for(int i = s.length() - 1; i >= 0; i--) {
      rev = rev + s.charAt(i);
  }
  

  → Loop starts from the last character and appends each character to rev. → At the end, rev contains the reversed string.

• Comparison:

  java

  if(s.equals(rev)) {
      System.out.println("This is palindrome string");
  } else {
      System.out.println("This is not palindrome string");
  }
  

  → If original string equals reversed string, it’s a palindrome.

🔹 Step 3: Correct Mistakes in Code (Interviewers notice this!)

• lenght() should be corrected to length().

• Method declaration should be:

  java

  public class PalindromeString {
      public static void main(String[] args) {
          //code
      }
  }
  

🔹 Step 4: Explain in Interview Style

Here’s a sample answer you can give:

“To check if a string is a palindrome, I reverse the string using a loop and compare it with the original. If both are equal, the string is a palindrome. For example, with i/p= MADAM, the o/p= MADAM, so the program prints that it is a palindrome. I used charAt() to fetch characters and equals() for comparison. Also, I ensured to use length() correctly to avoid compilation errors.”

-----------------------------------------------------------------------

//Method 2

//optimise code

You can mention optimization: Instead of reversing the whole string, compare characters from start and end moving inward (reduces time and memory). 

for(int i=0; i<s.length()/2; i++) {

    if(s.charAt(i) != s.charAt(s.length()-1-i)) {

        return false;

    }

}

return true;