StringBuilder and StringBuffer in Java - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new StringBuilder object.

Java

StringBuilder sb = new [1]();

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

AStringBuilder

BStringBuffer

CString

DStringBuilderBuffer

Attempts:

3 left

💡 Hint

Common Mistakes

Using String instead of StringBuilder, which is immutable.

Using a non-existent class name like StringBuilderBuffer.

The StringBuilder class is used to create mutable strings in Java. Here, we create a new instance using new StringBuilder().

2fill in blank

medium

Complete the code to append the string "Hello" to the StringBuffer object.

Java

StringBuffer sb = new StringBuffer();
sb.[1]("Hello");

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

Ainsert

Badd

Cconcat

Dappend

Attempts:

3 left

💡 Hint

Common Mistakes

Using add or concat which are not methods of StringBuffer.

Using insert which requires an index parameter.

The append method adds the given string to the end of the current StringBuffer content.

3fill in blank

hard

Fix the error in the code to convert StringBuilder to String.

Java

StringBuilder sb = new StringBuilder();
sb.append("Test");
String result = sb.[1]();

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

Astring

BtoString

Cconvert

DtoStr

Attempts:

3 left

💡 Hint

Common Mistakes

Using toStr or string which are not valid methods.

Trying to cast directly without toString().

The toString() method converts the StringBuilder content to a regular String.

4fill in blank

hard

Fill both blanks to create a StringBuffer, append "Hi", and convert to String.

Java

StringBuffer sb = new [1]();
sb.[2]("Hi");
String s = sb.toString();

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

AStringBuffer

BStringBuilder

Cappend

Dinsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using StringBuilder instead of StringBuffer for this task.

Using insert without an index parameter.

We create a StringBuffer object and use append to add text.

5fill in blank

hard

Fill all three blanks to create a StringBuilder, append "Bye", and convert to String.

Java

StringBuilder [1] = new StringBuilder();
[1].[2]("Bye");
String result = [1].[3]();

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

Asb

Bappend

CtoString

Dbuffer

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'buffer' as variable name which is not used here.

Using methods other than append or toString.

We name the StringBuilder sb, use append to add text, and toString() to get the String.

Practice

1.

Which of the following statements about StringBuilder and StringBuffer is true?

easy

A. StringBuffer is faster and not thread-safe, while StringBuilder is thread-safe but slower.

B. StringBuilder is faster but not thread-safe, while StringBuffer is thread-safe but slower.

C. Both StringBuilder and StringBuffer are thread-safe and have the same speed.

D. StringBuilder and StringBuffer cannot be used to modify strings.

Solution

1. Step 1: Understand thread safety difference

   StringBuffer is synchronized, making it thread-safe but slower.

2. Step 2: Understand performance difference

   StringBuilder is not synchronized, so it is faster but not thread-safe.

3. Final Answer:

   StringBuilder is faster but not thread-safe, while StringBuffer is thread-safe but slower. -> Option B

4. Quick Check:

   Thread safety and speed = A [OK]

Hint: Remember: Buffer is safe but slower; Builder is fast but unsafe [OK]

Common Mistakes:

• Confusing which class is thread-safe
• Thinking both have same speed
• Believing they cannot modify strings

2.

Which of the following is the correct way to create a StringBuilder object with initial content "Hello"?

easy

A. StringBuilder sb = new StringBuffer("Hello");

B. StringBuilder sb = StringBuilder("Hello");

C. StringBuilder sb = new StringBuilder(); sb.append("Hello");

D. StringBuilder sb = new StringBuilder("Hello");

Solution

1. Step 1: Check constructor syntax

   The correct way to create a StringBuilder with initial content is using new StringBuilder("Hello").

2. Step 2: Verify other options

   StringBuilder sb = StringBuilder("Hello"); misses new, StringBuilder sb = new StringBuilder(); sb.append("Hello"); is valid but not direct initialization, StringBuilder sb = new StringBuffer("Hello"); uses wrong class.

3. Final Answer:

   StringBuilder sb = new StringBuilder("Hello"); -> Option D

4. Quick Check:

   Correct constructor syntax = D [OK]

Hint: Use 'new' keyword and constructor with string inside parentheses [OK]

Common Mistakes:

• Omitting 'new' keyword
• Using StringBuffer instead of StringBuilder
• Trying to assign without constructor

3.

What is the output of the following code?

StringBuilder sb = new StringBuilder("Java");
sb.append(" Programming");
sb.replace(0, 4, "Python");
System.out.println(sb.toString());

medium

A. PythonJava Programming

B. Java Programming

C. Python Programming

D. ProgrammingPython

Solution

1. Step 1: Append " Programming" to "Java"

   After append, string becomes "Java Programming".

2. Step 2: Replace characters from index 0 to 4 with "Python"

   Characters "Java" (index 0 to 4) are replaced by "Python", resulting in "Python Programming".

3. Final Answer:

   Python Programming -> Option C

4. Quick Check:

   Append + replace result = A [OK]

Hint: Replace changes substring; append adds at end [OK]

Common Mistakes:

• Confusing replace indices
• Thinking append replaces text
• Ignoring toString() output

4.

Identify the error in the following code snippet:

StringBuffer sb = new StringBuffer("Test");
sb.insert(5, "ing");
System.out.println(sb);

medium

A. IndexOutOfBoundsException because insert index is invalid

B. NullPointerException because sb is not initialized

C. Compilation error due to wrong method name

D. No error; output is "Testing"

Solution

1. Step 1: Check insert index validity

   StringBuffer length is 4 ("Test"), but insert index is 5, which is out of bounds.

2. Step 2: Understand exception thrown

   Inserting at index greater than length throws IndexOutOfBoundsException.

3. Final Answer:

   IndexOutOfBoundsException because insert index is invalid -> Option A

4. Quick Check:

   Invalid insert index = B [OK]

Hint: Insert index must be between 0 and length inclusive [OK]

Common Mistakes:

• Assuming insert auto-corrects index
• Confusing insert with append
• Expecting no error on invalid index

5.

You want to build a thread-safe string that appends numbers 1 to 3 separated by commas, resulting in "1,2,3". Which code snippet correctly uses StringBuffer to achieve this?

1. StringBuffer sb = new StringBuffer();
   for (int i = 1; i <= 3; i++) {
       sb.append(i);
       if (i != 3) sb.append(",");
   }
   System.out.println(sb.toString());

2. StringBuffer sb = new StringBuffer();
   for (int i = 1; i <= 3; i++) {
       sb.append(i + ",");
   }
   System.out.println(sb.toString());

3. StringBuffer sb = new StringBuffer();
   sb.append("1,2,3,");
   System.out.println(sb.toString());

4. StringBuffer sb = new StringBuffer();
   for (int i = 1; i < 3; i++) {
       sb.append(i);
       sb.append(",");
   }
   System.out.println(sb.toString());

hard

A. Code snippet 1

B. Code snippet 2

C. Code snippet 3

D. Code snippet 4

Solution

1. Step 1: Analyze snippet 1

   Appends numbers 1 to 3 with commas except after last number, producing "1,2,3" correctly.

2. Step 2: Check other snippets

   Snippet 2 adds extra comma at end; snippet 3 hardcodes string but adds extra comma; snippet 4 misses last number 3.

3. Final Answer:

   Code snippet 1 -> Option A

4. Quick Check:

   Correct comma placement in loop = C [OK]

Hint: Add comma only if not last element in loop [OK]

Common Mistakes:

• Adding extra comma at end
• Missing last number in loop
• Hardcoding string without loop