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Javaprogramming~10 mins

String vs StringBuilder in Java - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
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'
AString
BStringBuilder
CStringBuffer
DStringBuilderBuilder
Attempts:
3 left
💡 Hint
Common Mistakes
Using String instead of StringBuilder causes immutability issues.
Trying to use a non-existent class like StringBuilderBuilder.
2fill in blank
medium

Complete the code to append text to a StringBuilder.

Java
sb.[1]("Hello");
Drag options to blanks, or click blank then click option'
Aadd
Bconcat
Cappend
Dinsert
Attempts:
3 left
💡 Hint
Common Mistakes
Using add which does not exist for StringBuilder.
Using concat which is a String method, not StringBuilder.
3fill in blank
hard

Fix the error in converting StringBuilder to String.

Java
String result = sb.[1]();
Drag options to blanks, or click blank then click option'
AtoString
BtoStr
Cstringify
Dconvert
Attempts:
3 left
💡 Hint
Common Mistakes
Using non-existent methods like toStr or stringify.
Trying to cast StringBuilder directly to String.
4fill in blank
hard

Fill both blanks to create a StringBuilder and append text.

Java
StringBuilder [1] = new [2]();
[1].append("World");
Drag options to blanks, or click blank then click option'
Asb
BStringBuilder
Cstr
DString
Attempts:
3 left
💡 Hint
Common Mistakes
Using String instead of StringBuilder for the object.
Using variable names that do not match the object usage.
5fill in blank
hard

Fill all three blanks to build a string and convert it to String.

Java
StringBuilder [1] = new [2]();
[1].append("Hi");
String [3] = [1].toString();
Drag options to blanks, or click blank then click option'
Abuilder
BStringBuilder
Ctext
DStringBuffer
Attempts:
3 left
💡 Hint
Common Mistakes
Using StringBuffer which is similar but different class.
Not converting StringBuilder to String before assignment.

Practice

(1/5)
1. Which statement best describes when to use StringBuilder instead of String in Java?
easy
A. Use StringBuilder when you need to change text many times efficiently.
B. Use StringBuilder for fixed text that never changes.
C. Use String when you want faster text modifications.
D. Use String to save memory when changing text repeatedly.

Solution

  1. Step 1: Understand the nature of String and StringBuilder

    String objects are immutable, meaning they cannot be changed once created. Changing text creates new objects, which is slow and uses more memory.
  2. Step 2: Identify when StringBuilder is better

    StringBuilder allows changing text without creating new objects, making it efficient for many modifications.
  3. Final Answer:

    Use StringBuilder when you need to change text many times efficiently. -> Option A
  4. Quick Check:

    Many changes = StringBuilder [OK]
Hint: Use StringBuilder for many text changes, String for fixed text [OK]
Common Mistakes:
  • Thinking String is better for many changes
  • Confusing immutability of String
  • Assuming StringBuilder uses more memory always
2. Which of the following is the correct way to create a StringBuilder object with initial text "Hello"?
easy
A. StringBuilder sb = StringBuilder("Hello");
B. StringBuilder sb = "Hello";
C. StringBuilder sb = new StringBuilder[] {"Hello"};
D. StringBuilder sb = new StringBuilder("Hello");

Solution

  1. Step 1: Recall the correct constructor syntax for StringBuilder

    The StringBuilder class requires the new keyword and parentheses with the initial string inside.
  2. Step 2: Check each option for syntax correctness

    StringBuilder sb = new StringBuilder("Hello"); uses new StringBuilder("Hello"), which is correct. Options A, B, and D have syntax errors or wrong usage.
  3. Final Answer:

    StringBuilder sb = new StringBuilder("Hello"); -> Option D
  4. Quick Check:

    Use new + constructor for StringBuilder [OK]
Hint: Always use 'new StringBuilder("text")' to create StringBuilder [OK]
Common Mistakes:
  • Omitting 'new' keyword
  • Using assignment without constructor
  • Using array syntax mistakenly
3. What is the output of this Java code?
StringBuilder sb = new StringBuilder("Hi");
sb.append("!");
sb.append(" How are you?");
System.out.println(sb.toString());
medium
A. Hi!How are you?
B. Hi How are you?
C. Hi! How are you?
D. Hi

Solution

  1. Step 1: Understand what append() does on StringBuilder

    Each append() adds the given text to the end of the current StringBuilder content without spaces unless explicitly added.
  2. Step 2: Trace the code execution

    Initial text is "Hi". First append adds "!" making "Hi!". Second append adds " How are you?" (with leading space), resulting in "Hi! How are you?".
  3. Final Answer:

    Hi! How are you? -> Option C
  4. Quick Check:

    Appends add text exactly as given [OK]
Hint: Append adds text exactly; watch spaces carefully [OK]
Common Mistakes:
  • Ignoring spaces in appended strings
  • Assuming append replaces text
  • Printing StringBuilder object directly without toString()
4. Identify the error in this code snippet:
StringBuilder sb = new StringBuilder("Test");
sb = sb + "ing";
System.out.println(sb);
medium
A. Cannot use + operator to concatenate StringBuilder and String.
B. Missing semicolon after StringBuilder initialization.
C. StringBuilder cannot be printed directly with println.
D. StringBuilder must be declared final.

Solution

  1. Step 1: Check the use of + operator with StringBuilder

    The + operator cannot be used to add a String to a StringBuilder object directly. This causes a compile-time error.
  2. Step 2: Understand correct way to append text to StringBuilder

    Use sb.append("ing") to add text instead of + operator.
  3. Final Answer:

    Cannot use + operator to concatenate StringBuilder and String. -> Option A
  4. Quick Check:

    Use append(), not +, with StringBuilder [OK]
Hint: Use append() to add text to StringBuilder, not + [OK]
Common Mistakes:
  • Trying to use + with StringBuilder
  • Thinking println can't print StringBuilder
  • Forgetting semicolons
5. You want to build a sentence by adding words one by one in a loop. Which approach is best for performance and why?
Option A:
String sentence = "";
for (String word : words) {
    sentence += word + " ";
}

Option B:
StringBuilder sentence = new StringBuilder();
for (String word : words) {
    sentence.append(word).append(" ");
}
hard
A. Option A is better because String is simpler to use.
B. Option B is better because StringBuilder avoids creating many new objects.
C. Both are equally efficient for building strings in loops.
D. Option A is better because it uses less memory.

Solution

  1. Step 1: Analyze Option B is better because String is simpler to use. (String concatenation in loop)

    Each concatenation with += creates a new String object because Strings are immutable, causing slow performance and more memory use.
  2. Step 2: Analyze Option B is better because StringBuilder avoids creating many new objects. (StringBuilder in loop)

    StringBuilder modifies the same object without creating new ones, making it faster and more memory efficient for repeated changes.
  3. Final Answer:

    Option B is better because StringBuilder avoids creating many new objects. -> Option B
  4. Quick Check:

    Use StringBuilder for repeated string changes in loops [OK]
Hint: Use StringBuilder in loops to build strings efficiently [OK]
Common Mistakes:
  • Using String concatenation in loops thinking it's fast
  • Ignoring immutability of String
  • Assuming simpler code is always better
Recommended

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10 questions~5 mineasy hard