Palindrome Checker

Check if a word or phrase reads the same forwards and backwards.

Difficulty

⭐ Beginner

Problem Description

Write a program that checks whether a given string is a palindrome. A palindrome is a word, phrase, or sequence that reads the same backwards as forwards.

Examples of palindromes:

• "racecar"
• "madam"
• "noon"
• "level"

Not palindromes:

• "hello"
• "world"
• "java"

Example Output

Input: "racecar"
Output: true

Input: "hello"
Output: false

Input: "noon"
Output: true

Skills Practised

• String manipulation
• Loops
• Conditional statements
• String methods (.length(), .charAt())

Requirements

1. Create a method isPalindrome(String text) that returns boolean
2. The method should return true if the text is a palindrome, false otherwise
3. For now, assume input is all lowercase with no spaces or punctuation

Test Cases

isPalindrome("racecar")  // should return true
isPalindrome("hello")    // should return false
isPalindrome("noon")     // should return true
isPalindrome("level")    // should return true
isPalindrome("java")     // should return false
isPalindrome("a")        // should return true
isPalindrome("")         // should return true

Hints

Hint 1: Reverse the string

One approach is to reverse the string and compare it with the original:

String reversed = "";
for (int i = text.length() - 1; i >= 0; i--) {
    reversed += text.charAt(i);
}
return text.equals(reversed);

Hint 2: Compare characters

Another approach is to compare characters from both ends:

for (int i = 0; i < text.length() / 2; i++) {
    if (text.charAt(i) != text.charAt(text.length() - 1 - i)) {
        return false;
    }
}
return true;

Hint 3: Two pointers

Use two pointers, one at the start and one at the end:

int left = 0;
int right = text.length() - 1;

while (left < right) {
    if (text.charAt(left) != text.charAt(right)) {
        return false;
    }
    left++;
    right--;
}
return true;

Starter Code

public class PalindromeChecker {
    public static void main(String[] args) {
        // Test your method
        System.out.println(isPalindrome("racecar"));  // true
        System.out.println(isPalindrome("hello"));    // false
        System.out.println(isPalindrome("noon"));     // true
    }

    public static boolean isPalindrome(String text) {
        // Write your code here
        return false;
    }
}

Solution

Click to reveal solution

Solution 1: Reverse and Compare

public class PalindromeChecker {
    public static void main(String[] args) {
        System.out.println(isPalindrome("racecar"));  // true
        System.out.println(isPalindrome("hello"));    // false
        System.out.println(isPalindrome("noon"));     // true
    }

    public static boolean isPalindrome(String text) {
        String reversed = "";

        for (int i = text.length() - 1; i >= 0; i--) {
            reversed += text.charAt(i);
        }

        return text.equals(reversed);
    }
}

Solution 2: Two Pointers (More Efficient)

public class PalindromeChecker {
    public static void main(String[] args) {
        System.out.println(isPalindrome("racecar"));  // true
        System.out.println(isPalindrome("hello"));    // false
        System.out.println(isPalindrome("noon"));     // true
    }

    public static boolean isPalindrome(String text) {
        int left = 0;
        int right = text.length() - 1;

        while (left < right) {
            if (text.charAt(left) != text.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }

        return true;
    }
}

Solution 3: Compare First and Second Half

public class PalindromeChecker {
    public static void main(String[] args) {
        System.out.println(isPalindrome("racecar"));  // true
        System.out.println(isPalindrome("hello"));    // false
        System.out.println(isPalindrome("noon"));     // true
    }

    public static boolean isPalindrome(String text) {
        for (int i = 0; i < text.length() / 2; i++) {
            if (text.charAt(i) != text.charAt(text.length() - 1 - i)) {
                return false;
            }
        }
        return true;
    }
}

Extension Challenges

1. Case Insensitive

Modify your solution to handle uppercase and lowercase:

isPalindrome("Racecar")  // should return true
isPalindrome("Level")    // should return true

Hint

Convert the string to lowercase first:

text = text.toLowerCase();

2. Ignore Spaces

Handle phrases with spaces:

isPalindrome("race car")  // should return true
isPalindrome("nurses run") // should return true

Hint

Remove spaces first:

text = text.replace(" ", "");

3. Full Phrase Palindrome

Handle complete phrases with spaces, punctuation, and mixed case:

isPalindrome("A man, a plan, a canal: Panama")  // should return true
isPalindrome("Was it a car or a cat I saw?")     // should return true

Hint

Keep only letters and numbers, and convert to lowercase:

text = text.replaceAll("[^a-zA-Z0-9]", "").toLowerCase();

4. Palindrome Number

Create a method that checks if a number is a palindrome:

isPalindromeNumber(121)    // should return true
isPalindromeNumber(123)    // should return false
isPalindromeNumber(12321)  // should return true

What You've Learned

After completing this challenge, you should understand:

• ✅ How to access individual characters in a string
• ✅ How to loop through a string
• ✅ How to compare characters
• ✅ Different approaches to solving the same problem
• ✅ The concept of "two pointers" technique

Related Concepts

This challenge introduces you to:

• String manipulation: Working with text data
• Algorithm efficiency: Some solutions are faster than others
• Edge cases: Empty strings, single characters

Next Challenge

Try the Sum of Digits challenge to practise working with numbers and loops!