String Basics and Memory Representation in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new String object from a character array.

DSA Java

char[] chars = {'H', 'e', 'l', 'l', 'o'};
String str = new String([1]);

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

Achars

Bstr

Cnew char[]

DString

Attempts:

3 left

💡 Hint

Common Mistakes

Using the String variable 'str' inside its own constructor.

Trying to pass 'new char[]' without specifying the array.

The String constructor takes a character array to create a new String object. Here, 'chars' is the character array.

2fill in blank

medium

Complete the code to compare two strings for content equality.

DSA Java

String a = "hello";
String b = new String("hello");
boolean result = a.[1](b);

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

A==

BcompareTo

Cequals

DequalsIgnoreCase

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' which compares references, not content.

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

The 'equals' method compares the content of two strings for equality.

3fill in blank

hard

Fix the error in the code to get the length of a string.

DSA Java

String s = "example";
int len = s.[1];

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

Alength

Blength()

Csize()

Dcount()

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'length' without parentheses which is for arrays, not strings.

Using 'size()' or 'count()' which are not String methods.

In Java, the length of a String is obtained by calling the method length() with parentheses.

4fill in blank

hard

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

DSA Java

String s = "programming";
String sub = s.substring([1], [2]);

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

A2

B5

C3

D6

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same value for start and end index.

Confusing inclusive and exclusive indices.

The substring method takes a start index (inclusive) and an end index (exclusive). To get characters at positions 2,3,4, use substring(2,5).

5fill in blank

hard

Fill all three blanks to create a map of words to their lengths for words longer than 3 characters.

DSA Java

Map<String, Integer> wordLengths = words.stream()
    .filter(word -> word.length() [1] 3)
    .collect(Collectors.toMap([2], [3]));

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

A>

Bword

CString::length

Dword -> word

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' in filter.

Using wrong lambda expressions in toMap.

The filter keeps words with length greater than 3. The toMap uses the word as key and its length as value.

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