String Traversal and Character Access in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print each character of the string using a for loop.

DSA Java

String word = "hello";
for (int i = 0; i [1] word.length(); i++) {
    System.out.print(word.charAt(i));
}

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

A<

B>=

C<=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes an error because it tries to access an index equal to word.length().

The loop should run from 0 up to but not including word.length() to avoid IndexOutOfBoundsException.

2fill in blank

medium

Complete the code to check if the first character of the string is 'A'.

DSA Java

String name = "Alice";
if (name.charAt([1]) == 'A') {
    System.out.println("Starts with A");
}

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

A1

B0

Cname.length()

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 accesses the second character, not the first.

The first character of a string is at index 0.

3fill in blank

hard

Fix the error in the code to print characters in reverse order.

DSA Java

String text = "code";
for (int i = text.length() - 1; i [1] 0; i--) {
    System.out.print(text.charAt(i));
}

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

A>=

B>

C<=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using > skips the character at index 0.

To include index 0, the loop condition should be i >= 0.

4fill in blank

hard

Fill both blanks to create a loop that prints only uppercase letters from the string.

DSA Java

String message = "HeLLo123";
for (int i = 0; i [1] message.length(); i++) {
    char ch = message.charAt(i);
    if (ch [2] 'A' && ch <= 'Z') {
        System.out.print(ch);
    }
}

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

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of >= skips the letter 'A'.

Using <= in loop condition causes index error.

The loop runs while i < message.length(). The condition checks if ch >= 'A' and ch <= 'Z' to find uppercase letters.

5fill in blank

hard

Fill all three blanks to create a loop that counts vowels in the string.

DSA Java

String input = "Education";
int count = 0;
for (int [1] = 0; i [2] input.length(); [3]++) {
    char c = input.charAt(i);
    if ("aeiouAEIOU".indexOf(c) != -1) {
        count++;
    }
}
System.out.println(count);

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

Ai

B<

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Using <= causes index error.

The loop variable is i, starting at 0, running while i < input.length(), and incrementing i each time.

Practice

1. What does the charAt() method do when used on a Java String?

easy

A. Returns the character at a specified index in the string

B. Returns the length of the string

C. Converts the string to uppercase

D. Splits the string into an array of substrings

Solution

1. Step 1: Understand the purpose of charAt()

   The charAt() method is designed to access a single character at a given index in a string.

2. Step 2: Compare with other string methods

   Methods like length() return the string length, toUpperCase() changes case, and split() divides the string. These are different from charAt().

3. Final Answer:

   Returns the character at a specified index in the string -> Option A

4. Quick Check:

   charAt() = character access [OK]

Hint: Remember: charAt gets one character by index [OK]

Common Mistakes:

• Confusing charAt() with length()
• Thinking charAt() splits the string
• Assuming charAt() changes string case

2. Which of the following is the correct way to loop through each character of a string str in Java?

easy

A. for (int i = 1; i < str.length(); i++) { char c = str.charAt(i); }

B. for (int i = 0; i <= str.length(); i++) { char c = str.charAt(i); }

C. for (int i = 0; i < str.length(); i++) { char c = str.charAt(i); }

D. for (int i = 1; i <= str.length(); i++) { char c = str.charAt(i); }

Solution

1. Step 1: Understand string indexing in Java

   String indices start at 0 and go up to length - 1. So loop must run from 0 to less than length.

2. Step 2: Check loop conditions

   for (int i = 0; i < str.length(); i++) { char c = str.charAt(i); } uses i < str.length(), which correctly stops before index equals length. Options B and D use <= which causes out-of-bounds error. Options C and D start at 1, skipping first character.

3. Final Answer:

   for (int i = 0; i < str.length(); i++) { char c = str.charAt(i); } -> Option C

4. Quick Check:

   Loop from 0 to length-1 [OK]

Hint: Loop from 0 to length-1 for string indices [OK]

Common Mistakes:

• Using <= instead of < causing index error
• Starting loop at 1 skipping first char
• Confusing string length with last index

3. What is the output of the following Java code?

String s = "hello";
for (int i = 0; i < s.length(); i++) {
  if (i % 2 == 0) {
    System.out.print(s.charAt(i));
  }
}

medium

A. hello

B. hlo

C. el

D. heo

Solution

1. Step 1: Identify loop and condition

   The loop runs from i=0 to 4 (length 5). It prints characters only when i is even (0,2,4).

2. Step 2: Find characters at even indices

   Index 0: 'h', Index 2: 'l', Index 4: 'o'. These characters are printed in order.

3. Final Answer:

   hlo -> Option B

4. Quick Check:

   Even indices chars = hlo [OK]

Hint: Print chars at even indices only [OK]

Common Mistakes:

• Printing all characters ignoring condition
• Confusing even and odd indices
• Off-by-one errors in loop

4. Consider this Java code snippet:

String text = "abc";
for (int i = 0; i <= text.length(); i++) {
  System.out.print(text.charAt(i));
}

What error will occur when running this code?

medium

A. StringIndexOutOfBoundsException

B. ArrayIndexOutOfBoundsException

C. NullPointerException

D. No error, prints 'abc'

Solution

1. Step 1: Analyze loop bounds

   The loop runs from i=0 to i=3 (because of <= and length=3). Valid indices are 0,1,2 only.

2. Step 2: Identify error cause

   At i=3, charAt(3) is called, which is out of string bounds, causing StringIndexOutOfBoundsException.

3. Final Answer:

   StringIndexOutOfBoundsException -> Option A

4. Quick Check:

   charAt index out of range causes StringIndexOutOfBoundsException [OK]

Hint: Loop must use < length, not <= length [OK]

Common Mistakes:

• Confusing StringIndexOutOfBoundsException with ArrayIndexOutOfBoundsException
• Assuming no error occurs
• Thinking NullPointerException will happen

5. Given the string input = "a1b2c3d4", which code snippet correctly extracts and prints only the digits in order?

hard

A. for (int i = 0; i < input.length(); i++) { if (Character.isLetter(input.charAt(i))) System.out.print(input.charAt(i)); }

B. for (int i = 0; i < input.length(); i++) { if (input.charAt(i) >= 'a' && input.charAt(i) <= 'z') System.out.print(input.charAt(i)); }

C. for (int i = 0; i < input.length(); i++) { System.out.print(input.charAt(i)); }

D. for (int i = 0; i < input.length(); i++) { if (input.charAt(i) >= '0' && input.charAt(i) <= '9') System.out.print(input.charAt(i)); }

Solution

1. Step 1: Understand the goal

   We want to print only digits from the string, ignoring letters.

2. Step 2: Check each option's condition

   for (int i = 0; i < input.length(); i++) { if (Character.isLetter(input.charAt(i))) System.out.print(input.charAt(i)); } uses Character.isLetter(), prints letters. for (int i = 0; i < input.length(); i++) { if (input.charAt(i) >= 'a' && input.charAt(i) <= 'z') System.out.print(input.charAt(i)); } checks lowercase letters range, prints letters. for (int i = 0; i < input.length(); i++) { System.out.print(input.charAt(i)); } prints all characters. for (int i = 0; i < input.length(); i++) { if (input.charAt(i) >= '0' && input.charAt(i) <= '9') System.out.print(input.charAt(i)); } checks if char between '0' and '9', prints "1234" correctly.

3. Final Answer:

   for (int i = 0; i < input.length(); i++) { if (input.charAt(i) >= '0' && input.charAt(i) <= '9') System.out.print(input.charAt(i)); } -> Option D

4. Quick Check:

   Check char between '0' and '9' to find digits [OK]

Hint: Check if char is between '0' and '9' for digits [OK]

Common Mistakes:

• Printing letters instead of digits
• Printing all characters without filtering
• Using wrong character range for digits