String Basics and Memory Representation in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of String Concatenation and Memory

What is the output of the following Java code snippet?

DSA Java

String a = "Hello";
String b = "World";
String c = a + b;
System.out.println(c);

A "HelloWorld "

B "Hello World"

C "Hello+World"

D "HelloWorld"

💡 Hint

Remember that + operator concatenates strings without adding spaces automatically.

The + operator joins the two strings exactly as they are, so "Hello" + "World" becomes "HelloWorld" without spaces.

❓ Predict Output

intermediate

String Immutability Effect

What will be printed after running this Java code?

DSA Java

String s = "abc";
s.toUpperCase();
System.out.println(s);

A "abc"

B null

C "Abc"

D "ABC"

💡 Hint

Strings in Java cannot be changed once created.

The toUpperCase() method returns a new string but does not change the original string s.

🔧 Debug

advanced

Why does this String comparison fail?

Consider this Java code snippet. Why does it print "Not equal"?

DSA Java

String s1 = new String("test");
String s2 = new String("test");
if (s1 == s2) {
    System.out.println("Equal");
} else {
    System.out.println("Not equal");
}

A Because new String() creates the same object, but == compares content incorrectly.

B Because == compares references, not content, so s1 and s2 are different objects.

C Because s1 and s2 are interned strings and == always returns false for interned strings.

D Because the strings have different content.

💡 Hint

Use .equals() to compare string content in Java.

The == operator checks if both variables point to the same object in memory. Here, s1 and s2 are different objects with the same content.

🧠 Conceptual

advanced

String Pool Behavior

Which statement about Java String pool is true?

A Strings created with new keyword are always stored in the String pool.

B Every new String object is stored in the String pool automatically.

C String literals are stored in the String pool and reused to save memory.

D String pool stores mutable strings to allow changes.

💡 Hint

Think about how Java optimizes memory for string literals.

Java stores string literals in a special pool to reuse them and save memory. Strings created with new keyword are not automatically pooled.

❓ Predict Output

expert

Output of String Interning

What is the output of this Java code snippet?

DSA Java

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

A

false
true

B

true
true

C

false
false

D

true
false

💡 Hint

intern() returns a reference from the String pool.

s1 is a new object, so s1 == s2 is false. s2 and s3 both refer to the pooled string literal, so s2 == s3 is true.

Practice

1. What is the correct way to create a string literal in Java?

easy

A. String s = 'Hello';

B. String s = new String();

C. String s = "Hello";

D. String s = String();

Solution

1. Step 1: Understand string literal syntax

   In Java, string literals are enclosed in double quotes, so "Hello" is correct.

2. Step 2: Check each option

   String s = "Hello"; uses double quotes correctly. String s = new String(); creates an empty string object but not a literal. String s = 'Hello'; uses single quotes which are for characters, not strings. String s = String(); is invalid syntax.

3. Final Answer:

   String s = "Hello"; -> Option C

4. Quick Check:

   String literal uses double quotes = A [OK]

Hint: Use double quotes for string literals in Java [OK]

Common Mistakes:

• Using single quotes for strings
• Trying to call String() like a function
• Confusing new String() with literal

2. Which of the following is the correct syntax to compare two strings s1 and s2 for content equality in Java?

easy

A. if (s1.equals(s2))

B. if (s1 == s2)

C. if (s1 = s2)

D. if (s1.compare(s2))

Solution

1. Step 1: Understand string comparison in Java

   Using == compares references (memory addresses), not content. To compare content, use the equals() method.

2. Step 2: Evaluate each option

   if (s1 == s2) compares references, not content. if (s1.equals(s2)) correctly uses equals() method. if (s1 = s2) uses assignment (=) which is invalid in condition. if (s1.compare(s2)) uses a non-existent method compare().

3. Final Answer:

   if (s1.equals(s2)) -> Option A

4. Quick Check:

   Use equals() for string content comparison = B [OK]

Hint: Use equals() method to compare string contents [OK]

Common Mistakes:

• Using == to compare strings
• Using assignment operator in if condition
• Calling non-existent methods

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

String s1 = "hello";
String s2 = "hello";
String s3 = new String("hello");
System.out.println(s1 == s2);
System.out.println(s1 == s3);
System.out.println(s1.equals(s3));

medium

A. true
false
true

B. false
false
true

C. true
true
true

D. false
true
false

Solution

1. Step 1: Understand string pool and new String()

   s1 and s2 are string literals and point to the same object in the string pool, so s1 == s2 is true. s3 is a new String object, so s1 == s3 is false because they reference different objects.

2. Step 2: Check equals() method

   s1.equals(s3) compares content and both have "hello", so it returns true.

3. Final Answer:

   true false true -> Option A

4. Quick Check:

   == compares references, equals() compares content = C [OK]

Hint: == checks memory, equals() checks string content [OK]

Common Mistakes:

• Assuming new String() shares reference
• Confusing == with equals()
• Ignoring string pool behavior

4. Identify the error in the following code snippet:

String s = "test";
if (s == "test") {
    System.out.println("Match");
} else {
    System.out.println("No Match");
}

medium

A. System.out.println syntax is incorrect

B. Missing semicolon after string declaration

C. String literals cannot be compared

D. Using == to compare strings can cause unexpected results

Solution

1. Step 1: Check string comparison method

   The code uses == to compare strings, which compares references, not content. This can cause unexpected results if strings are not from the pool.

2. Step 2: Verify other syntax

   There is no missing semicolon, string literals can be compared with equals(), and System.out.println syntax is correct.

3. Final Answer:

   Using == to compare strings can cause unexpected results -> Option D

4. Quick Check:

   Use equals() for string comparison, not == = A [OK]

Hint: Avoid == for strings; use equals() instead [OK]

Common Mistakes:

• Thinking == compares string content
• Ignoring string pool behavior
• Assuming syntax errors where none exist

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

String a = "Java";
String b = "Ja" + "va";
String c = new String("Java");
System.out.println(a == b);
System.out.println(a == c);
System.out.println(a.equals(c));

Explain why the outputs are as they are.

hard

A. false
true
false

B. true
false
true

C. true
true
true

D. false
false
true

Solution

1. Step 1: Analyze string concatenation with literals

   Strings "Ja" and "va" are literals concatenated at compile time, so b points to the same string pool object as a. Therefore, a == b is true.

2. Step 2: Understand new String() behavior

   c is a new String object created at runtime, so a == c is false because they reference different objects. However, a.equals(c) compares content and returns true.

3. Final Answer:

   true false true -> Option B

4. Quick Check:

   Compile-time concat shares pool, new String() creates new object = D [OK]

Hint: Compile-time concat shares pool; new String() creates new object [OK]

Common Mistakes:

• Assuming runtime concat shares pool
• Confusing == with equals()
• Ignoring compile-time optimization