How Strings Work Differently Across Languages in DSA Java - Code Step by Step

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new String object in Java.

DSA Java

String greeting = new [1]("Hello");

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

Achar

BString

Cstring

DText

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase 'string' instead of 'String'.

Using 'char' which is for single characters.

In Java, the class name String must be capitalized to create a new String object.

2fill in blank

medium

Complete the code to compare two strings for equality in Java.

DSA Java

if (str1.[1](str2)) {
    System.out.println("Equal");
}

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

Aequals

B==

CcompareTo

DequalsIgnoreCase

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' to compare strings.

Using 'compareTo' which returns an int, not boolean.

Use equals() method to compare string contents in Java.

3fill in blank

hard

Fix the error in the code to concatenate two strings correctly.

DSA Java

String fullName = firstName [1] lastName;

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

Aconcat()

B&

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '&' which is not valid for string concatenation in Java.

Using '-' which subtracts numbers, not concatenate strings.

In Java, the '+' operator concatenates strings.

4fill in blank

hard

Fill both blanks to create a substring from index 2 to 5.

DSA Java

String part = text.[1]([2], 6);

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

Asubstring

Bsubstr

C2

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'substr' which is not a Java String method.

Using wrong start index.

Java uses substring(startIndex, endIndex) to get part of a string. Index 2 is the start.

5fill in blank

hard

Fill all three blanks to convert a string to uppercase and check if it equals "HELLO".

DSA Java

if (input.[1]().[2]("HELLO") [3] true) {
    System.out.println("Match");
}

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

AtoUpperCase

Bequals

C==

DequalsIgnoreCase

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'equalsIgnoreCase' after converting to uppercase (redundant).

Using '==' to compare strings directly.

Convert string to uppercase with toUpperCase(), then compare with equals(). The '==' compares boolean true.

Practice

1. In Java, which of the following statements about String objects is true?

easy

A. Strings are stored as linked lists of characters.

B. Strings can be changed directly by modifying their characters.

C. Strings are immutable, meaning once created, they cannot be changed.

D. Strings are always stored in the stack memory.

Solution

1. Step 1: Understand Java String immutability

   In Java, String objects cannot be changed after creation. Any modification creates a new String object.

2. Step 2: Compare other options

   Strings are not linked lists, and they are stored in heap memory, not stack.

3. Final Answer:

   Strings are immutable, meaning once created, they cannot be changed. -> Option C

4. Quick Check:

   Java String immutability = true [OK]

Hint: Remember: Java Strings never change after creation [OK]

Common Mistakes:

• Thinking Strings can be changed in place
• Confusing String storage with primitive types
• Assuming Strings are linked lists

2. Which of the following is the correct way to create a new String object in Java?

easy

A. String s = new String("hello");

B. String s = String("hello");

C. String s = new string("hello");

D. String s = 'hello';

Solution

1. Step 1: Check Java String creation syntax

   The correct syntax uses the keyword new and capitalized String class: new String("hello").

2. Step 2: Identify incorrect options

   String s = String("hello"); misses new, C uses lowercase string, and D uses single quotes which are for chars, not Strings.

3. Final Answer:

   String s = new String("hello"); -> Option A

4. Quick Check:

   Correct String constructor syntax = String s = new String("hello"); [OK]

Hint: Use new String("text") with capital S and double quotes [OK]

Common Mistakes:

• Using single quotes for Strings
• Using lowercase 'string' instead of 'String'
• Omitting the new keyword

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

String a = "hello";
String b = "hello";
System.out.println(a == b);

medium

A. Runtime exception

B. false

C. Compilation error

D. true

Solution

1. Step 1: Understand String literals and pooling in Java

   Both a and b refer to the same String literal in the String pool, so a == b compares the same object reference.

2. Step 2: Evaluate the == operator on String literals

   Since both point to the same object, a == b returns true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   String literal pooling makes == true [OK]

Hint: String literals share references, so == returns true [OK]

Common Mistakes:

• Confusing == with .equals() for content check
• Assuming == always returns false for Strings
• Ignoring String pool behavior

4. Identify the error in the following Java code snippet:

String s = "hello";
s[0] = 'H';
System.out.println(s);

medium

A. Strings are immutable; characters cannot be changed by index.

B. Syntax error: single quotes cannot be used for characters.

C. Runtime error: index out of bounds.

D. No error; output will be "Hello".

Solution

1. Step 1: Check String immutability in Java

   Strings cannot be changed by modifying characters at an index because they are immutable.

2. Step 2: Analyze the code error

   Attempting s[0] = 'H'; causes a compile-time error since Strings do not support indexing for assignment.

3. Final Answer:

   Strings are immutable; characters cannot be changed by index. -> Option A

4. Quick Check:

   Immutable Strings prevent index assignment [OK]

Hint: Strings can't change characters by index in Java [OK]

Common Mistakes:

• Trying to modify String characters directly
• Confusing char arrays with Strings
• Expecting runtime error instead of compile error

5. Given the following Java code, what will be the output?

String s1 = new String("test");
String s2 = "test";
String s3 = s1.intern();
System.out.println(s1 == s2);
System.out.println(s2 == s3);

hard

A. true\nfalse

B. false\ntrue

C. false\nfalse

D. true\ntrue

Solution

1. Step 1: Understand new String() vs String literal

   s1 is a new object in heap, s2 is a String literal from the pool, so s1 == s2 is false.

2. Step 2: Understand intern() method

   s3 is the interned version of s1, which returns the pooled String equal to "test", same as s2. So s2 == s3 is true.

3. Final Answer:

   false true -> Option B

4. Quick Check:

   new String != literal, intern() returns pooled String [OK]

Hint: intern() returns pooled String; new String creates new object [OK]

Common Mistakes:

• Assuming new String equals literal by ==
• Ignoring intern() effect on references
• Confusing == with .equals()