Why Strings Are a Data Structure Not Just Text in DSA Java - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a string variable named greeting with the value "Hello".

DSA Java

String greeting = [1];

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

Agreeting

BHello

C'Hello'

D"Hello"

Attempts:

3 left

💡 Hint

Common Mistakes

Using single quotes instead of double quotes for strings.

Not using quotes at all, which causes a syntax error.

In Java, strings must be enclosed in double quotes to be valid string literals.

2fill in blank

medium

Complete the code to find the length of the string stored in variable word.

DSA Java

int length = word.[1]();

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

Alength

Bsize

Ccount

DgetLength

Attempts:

3 left

💡 Hint

Common Mistakes

Using size() which is for collections, not strings.

Using count() or getLength() which are not valid string methods.

The length() method returns the number of characters in a Java string.

3fill in blank

hard

Fix the error in the code to get the character at index 2 from the string text.

DSA Java

char ch = text.[1](2);

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

AgetChar

BcharAt

Cchar

Dat

Attempts:

3 left

💡 Hint

Common Mistakes

Using getChar or at which are not valid string methods.

Trying to access characters like an array without a method.

The correct method to get a character at a specific index in a string is charAt(index).

4fill in blank

hard

Fill both blanks to create a substring from index 1 to 4 of the string s.

DSA Java

String sub = s.[1]([2], 5);

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

Asubstring

Bsubstr

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using substr which is not a Java String method.

Using wrong start index like 0 instead of 1.

The substring method takes a start index and an end index (exclusive). Here, start is 1 and end is 5.

5fill in blank

hard

Fill all three blanks to check if the string str contains the word "data" ignoring case.

DSA Java

boolean containsData = str.[1]().[2]("data");
if (containsData [3] true) {
    System.out.println("Found data!");
}

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

AtoLowerCase

Bcontains

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of == in the if condition.

Not converting string to lower case before contains check.

Convert string to lower case, then check if it contains "data". Use == to compare boolean true.

Practice

1. Why are strings considered a data structure and not just plain text in programming?

easy

A. Because strings store characters in a sequence allowing indexing and manipulation

B. Because strings are only used for displaying text on screen

C. Because strings cannot be changed once created

D. Because strings only hold numbers and not letters

Solution

1. Step 1: Understand what a data structure is

   A data structure organizes data so it can be accessed and changed efficiently.

2. Step 2: Analyze string properties

   Strings store characters in order, allowing access by position and operations like slicing or concatenation.

3. Final Answer:

   Because strings store characters in a sequence allowing indexing and manipulation -> Option A

4. Quick Check:

   Strings = sequence + access [OK]

Hint: Strings hold ordered characters, not just text display [OK]

Common Mistakes:

• Thinking strings are only for display
• Believing strings cannot be accessed by index
• Confusing strings with numbers only

2. Which of the following is the correct way to declare a string in Java?

easy

A. String s = 'hello';

B. String s = "hello";

C. string s = "hello";

D. String s = hello;

Solution

1. Step 1: Recall Java string declaration syntax

   Java strings use the class String and double quotes for text.

2. Step 2: Check each option

   String s = "hello"; uses String with double quotes correctly; others have syntax errors or wrong case.

3. Final Answer:

   String s = "hello"; -> Option B

4. Quick Check:

   Java strings = String + double quotes [OK]

Hint: Use double quotes and capital S for String in Java [OK]

Common Mistakes:

• Using single quotes for strings
• Using lowercase 'string' instead of 'String'
• Not using quotes around text

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

String s = "hello";
System.out.println(s.charAt(1));

medium

A. e

B. l

C. h

D. o

Solution

1. Step 1: Understand charAt method

   charAt(index) returns the character at the given zero-based index.

2. Step 2: Find character at index 1 in "hello"

   Index 0 = 'h', index 1 = 'e', so output is 'e'.

3. Final Answer:

   e -> Option A

4. Quick Check:

   charAt(1) = 'e' [OK]

Hint: Remember string indexing starts at 0 in Java [OK]

Common Mistakes:

• Thinking indexing starts at 1
• Confusing charAt with substring
• Printing the whole string instead of one char

4. Identify the error in this Java code snippet:

String s = "data";
s.charAt(4);

medium

A. No error, code runs fine

B. Syntax error due to missing semicolon

C. IndexOutOfBoundsException because index 4 is invalid

D. NullPointerException because s is null

Solution

1. Step 1: Check string length and valid indices

   "data" length is 4, valid indices are 0 to 3.

2. Step 2: Analyze charAt(4) call

   Index 4 is out of range, causing IndexOutOfBoundsException at runtime.

3. Final Answer:

   IndexOutOfBoundsException because index 4 is invalid -> Option C

4. Quick Check:

   charAt index must be less than length [OK]

Hint: Index must be less than string length [OK]

Common Mistakes:

• Assuming index 4 is valid for length 4
• Confusing syntax errors with runtime errors
• Thinking string indices start at 1

5. Given the string String s = "abcabc";, which code snippet correctly counts the number of times the character 'a' appears?

hard

A. int count = 0;
for(int i=1; i

B. int count = 0;
for(char c : s.toCharArray()) {
if(c == 'A') count++;
}
System.out.println(count);

C. int count = s.indexOf('a');
System.out.println(count);

D. int count = s.length() - s.replace("a", "").length();
System.out.println(count);

Solution

1. Step 1: Verify the replace trick

   For s="abcabc", s.replace("a","")="bcbc" (length 4), 6-4=2 correct.

2. Step 2: Why others fail

   Char array loop checking uppercase 'A' finds zero; indexOf('a') returns first position 0; index loop starting at 1 misses first 'a' at 0, finds only one.

3. Final Answer:

   int count = s.length() - s.replace("a", "").length(); System.out.println(count); -> Option D

4. Quick Check:

   Count 'a' = original length - length without 'a' [OK]

Hint: Use length difference after replace to count chars [OK]

Common Mistakes:

• Using indexOf to count occurrences (gives position)
• Case mismatch: checking uppercase 'A'
• Off-by-one: starting loop index from 1