Palindrome Detection in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Palindrome Check for a Simple String

What is the output of the following Java code that checks if a string is a palindrome?

DSA Java

public class Palindrome {
    public static boolean isPalindrome(String s) {
        int left = 0, right = s.length() - 1;
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }

    public static void main(String[] args) {
        System.out.println(isPalindrome("radar"));
    }
}

A true

B false

C Compilation error

D Runtime exception

💡 Hint

Think about how the characters at the start and end of the string compare.

The string "radar" reads the same forwards and backwards, so the method returns true.

❓ Predict Output

intermediate

Output of Palindrome Check with Mixed Case

What is the output of this Java code checking palindrome for a mixed case string?

DSA Java

public class Palindrome {
    public static boolean isPalindrome(String s) {
        int left = 0, right = s.length() - 1;
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }

    public static void main(String[] args) {
        System.out.println(isPalindrome("Racecar"));
    }
}

A false

B Runtime exception

C Compilation error

D true

💡 Hint

Remember that character comparison is case sensitive.

The method compares characters exactly, so 'R' and 'r' are different, causing false.

🔧 Debug

advanced

Identify the Error in Palindrome Check Code

What error will this Java code produce when checking palindrome?

DSA Java

public class Palindrome {
    public static boolean isPalindrome(String s) {
        int left = 0, right = s.length();
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }

    public static void main(String[] args) {
        System.out.println(isPalindrome("level"));
    }
}

A false

B StringIndexOutOfBoundsException

C true

D Compilation error

💡 Hint

Check how the right index is initialized and used.

The right index is set to s.length(), which is out of bounds for charAt, causing an exception.

🧠 Conceptual

advanced

Understanding Palindrome Detection with Ignored Characters

Which approach correctly detects palindromes ignoring spaces and punctuation in Java?

A Reverse the string and compare without ignoring any characters.

B Check palindrome directly without any string modification.

C Only convert to lowercase but keep spaces and punctuation.

D Remove non-alphanumeric characters and convert to lowercase before checking palindrome.

💡 Hint

Think about how spaces and punctuation affect palindrome checks.

Ignoring non-alphanumeric characters and case ensures correct palindrome detection for phrases.

🚀 Application

expert

Output of Palindrome Check for Linked List

Given a singly linked list representing characters, what is the output after checking if it forms a palindrome?

DSA Java

class Node {
    char val;
    Node next;
    Node(char val) { this.val = val; this.next = null; }
}

public class PalindromeLinkedList {
    public static boolean isPalindrome(Node head) {
        if (head == null) return true;
        Node slow = head, fast = head;
        java.util.Stack<Character> stack = new java.util.Stack<>();
        while (fast != null && fast.next != null) {
            stack.push(slow.val);
            slow = slow.next;
            fast = fast.next.next;
        }
        if (fast != null) slow = slow.next;
        while (slow != null) {
            if (stack.pop() != slow.val) return false;
            slow = slow.next;
        }
        return true;
    }

    public static void main(String[] args) {
        Node head = new Node('a');
        head.next = new Node('b');
        head.next.next = new Node('b');
        head.next.next.next = new Node('a');
        System.out.println(isPalindrome(head));
    }
}

A false

B Compilation error

C true

D Runtime exception

💡 Hint

Think about how the stack helps compare the first half with the second half.

The linked list 'a'->'b'->'b'->'a' reads the same forwards and backwards, so returns true.

Practice

1. What does a palindrome string mean in the context of palindrome detection?

easy

A. A string with all unique characters

B. A string that contains only vowels

C. A string that is sorted alphabetically

D. A string that reads the same forwards and backwards

Solution

1. Step 1: Understand palindrome definition

   A palindrome is a word or phrase that reads the same forwards and backwards.

2. Step 2: Compare options with definition

   Only "a string that reads the same forwards and backwards" matches this definition exactly.

3. Final Answer:

   A string that reads the same forwards and backwards -> Option D

4. Quick Check:

   Palindrome = Same forwards and backwards [OK]

Hint: Palindrome means mirror reading same both ways [OK]

Common Mistakes:

• Confusing palindrome with unique characters
• Thinking palindrome means sorted string
• Assuming palindrome only has vowels

2. Which of the following Java code snippets correctly checks if a string s is a palindrome?

easy

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

B. for (int i = 0; i < s.length(); i++) { if (s.charAt(i) != s.charAt(i + 1)) return false; } return true;

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

D. for (int i = 0; i < s.length(); i++) { if (s.charAt(i) != s.charAt(s.length() - i)) return false; } return true;

Solution

1. Step 1: Check loop and character comparison logic

   for (int i = 0; i < s.length() / 2; i++) { if (s.charAt(i) != s.charAt(s.length() - 1 - i)) return false; } return true; compares characters from start and end moving towards center correctly.

2. Step 2: Identify errors in other options

   Adjacent chars comparison is incorrect; equality check instead of inequality is wrong; off-by-one index error.

3. Final Answer:

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

4. Quick Check:

   Compare chars from ends moving inward [OK]

Hint: Compare chars from start and end moving inward [OK]

Common Mistakes:

• Using wrong indices causing index out of bounds
• Comparing adjacent characters instead of mirrored ones
• Incorrect loop range causing incomplete checks

3. What will be the output of the following Java code?

String s = "racecar";
boolean isPalindrome = true;
for (int i = 0; i < s.length() / 2; i++) {
    if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
        isPalindrome = false;
        break;
    }
}
System.out.println(isPalindrome);

medium

A. false

B. Compilation error

C. true

D. Runtime exception

Solution

1. Step 1: Analyze the string and loop logic

   String "racecar" is a palindrome. Loop compares mirrored characters correctly.

2. Step 2: Check if any mismatch occurs

   No mismatch found, so isPalindrome remains true and prints true.

3. Final Answer:

   true -> Option C

4. Quick Check:

   "racecar" reads same forwards and backwards [OK]

Hint: Check mirrored chars; no mismatch means true [OK]

Common Mistakes:

• Assuming palindrome check fails for odd length strings
• Confusing loop range causing missed checks
• Misreading output as string instead of boolean

4. Identify the error in the following Java code that attempts to check if a string str is a palindrome:

boolean checkPalindrome(String str) {
    for (int i = 0; i <= str.length() / 2; i++) {
        if (str.charAt(i) != str.charAt(str.length() - i)) {
            return false;
        }
    }
    return true;
}

medium

A. The loop condition should be i < str.length() / 2 instead of <=.

B. The index str.length() - i causes an index out of bounds error.

C. The method should return void instead of boolean.

D. The comparison operator should be == instead of !=.

Solution

1. Step 1: Check index usage in charAt

   Index str.length() - i is off by one; last valid index is str.length() - 1.

2. Step 2: Identify error cause

   Accessing str.charAt(str.length() - i) when i=0 causes index out of bounds.

3. Final Answer:

   Index str.length() - i causes out of bounds error -> Option B

4. Quick Check:

   String indices start at 0, max index = length-1 [OK]

Hint: Check if indices go beyond string length - 1 [OK]

Common Mistakes:

• Using <= in loop causing extra iteration
• Confusing comparison operators
• Wrong return type for boolean method

5. Given a list of strings ["level", "world", "deed", "java", "noon"], which Java code snippet correctly returns a new list containing only the palindromes?

hard

A. List<String> palindromes = list.stream().filter(s -> new StringBuilder(s).reverse().toString().equals(s)).collect(Collectors.toList());

B. List<String> palindromes = list.stream().filter(s -> s.charAt(0) == s.charAt(s.length() - 1)).collect(Collectors.toList());

C. List<String> palindromes = list.stream().filter(s -> s.length() % 2 == 0).collect(Collectors.toList());

D. List<String> palindromes = list.stream().filter(s -> s.equals(s.toUpperCase())).collect(Collectors.toList());

Solution

1. Step 1: Understand palindrome filtering logic

   List<String> palindromes = list.stream().filter(s -> new StringBuilder(s).reverse().toString().equals(s)).collect(Collectors.toList()); reverses string and compares to original, correctly identifying palindromes.

2. Step 2: Analyze other options

   Checking only first and last char equality; filtering even length strings; checking uppercase equality; all incorrect.

3. Final Answer:

   List<String> palindromes = list.stream().filter(s -> new StringBuilder(s).reverse().toString().equals(s)).collect(Collectors.toList()); -> Option A

4. Quick Check:

   Reverse string equals original means palindrome [OK]

Hint: Reverse string and compare to original to find palindrome [OK]

Common Mistakes:

• Checking only first and last characters
• Filtering by string length instead of palindrome
• Confusing uppercase check with palindrome