Palindrome Detection in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a string is a palindrome by reversing it.

DSA Java

String reversed = new StringBuilder(input).[1]().toString();

Drag options to blanks, or click blank then click option'

Atrim

BtoUpperCase

Csubstring

Dreverse

Attempts:

3 left

💡 Hint

Common Mistakes

Using toUpperCase() instead of reverse()

Trying substring() which extracts part of string

Using trim() which removes spaces

The reverse() method of StringBuilder reverses the characters, which is needed to check palindrome.

2fill in blank

medium

Complete the code to compare the original and reversed strings ignoring case.

DSA Java

boolean isPalindrome = input.[1](reversed);

Drag options to blanks, or click blank then click option'

Aequals

BequalsIgnoreCase

CcompareTo

Dcontains

Attempts:

3 left

💡 Hint

Common Mistakes

Using equals() which is case sensitive

Using compareTo() which returns int, not boolean

Using contains() which checks substring presence

equalsIgnoreCase compares two strings ignoring case differences, useful for palindrome check.

3fill in blank

hard

Fix the error in the loop that checks characters from start and end for palindrome.

DSA Java

for (int i = 0; i < input.length() / [1]; i++) {
    if (input.charAt(i) != input.charAt(input.length() - 1 - i)) {
        return false;
    }
}

Drag options to blanks, or click blank then click option'

A1

B4

C2

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Checking full length causing index errors

Dividing by 3 or 4 which misses some pairs

Dividing by 1 which does nothing

We only need to check up to half the string length, so dividing by 2 is correct.

4fill in blank

hard

Fill both blanks to create a map of words to their palindrome status.

DSA Java

Map<String, Boolean> palindromeMap = words.stream()
    .collect(Collectors.toMap(word -> word[1], word -> isPalindrome(word) [2]));

Drag options to blanks, or click blank then click option'

A.toLowerCase()

B== true

C!= false

D.trim()

Attempts:

3 left

💡 Hint

Common Mistakes

Using '== true' which is redundant

Not normalizing keys causing duplicates

Using .trim() which doesn't normalize case

We normalize words to lowercase as keys and check if isPalindrome(word) is not false for values.

5fill in blank

hard

Fill all three blanks to filter and collect only palindrome words in uppercase.

DSA Java

List<String> palindromes = words.stream()
    .filter(word -> word.[1]())
    .map(word -> word.[2]())
    .collect(Collectors.[3]());

Drag options to blanks, or click blank then click option'

AequalsIgnoreCase(new StringBuilder(word).reverse().toString)

BtoUpperCase

CtoList

DequalsIgnoreCase(new StringBuilder(word).reverse().toString())

Attempts:

3 left

💡 Hint

Common Mistakes

Missing parentheses in toString() causing errors

Using equalsIgnoreCase without reverse string

Using collect(Collectors.toSet()) instead of toList()

Filter uses equalsIgnoreCase comparing word to its reversed string, map converts to uppercase, collect gathers to list.

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