String Traversal and Character Access in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of character traversal with index increment

What is the output of the following Java code snippet?

DSA Java

public class Main {
    public static void main(String[] args) {
        String s = "hello";
        StringBuilder result = new StringBuilder();
        for (int i = 0; i < s.length(); i += 2) {
            result.append(s.charAt(i));
        }
        System.out.println(result.toString());
    }
}

A hlo

B helo

C el

D heo

💡 Hint

Look at how the loop increments the index by 2 each time.

The loop starts at index 0 and increments by 2, so it accesses characters at positions 0, 2, and 4 which are 'h', 'l', and 'o'.

❓ Predict Output

intermediate

Output of reverse string traversal

What will be printed by this Java code?

DSA Java

public class Main {
    public static void main(String[] args) {
        String s = "world";
        StringBuilder reversed = new StringBuilder();
        for (int i = s.length() - 1; i >= 0; i--) {
            reversed.append(s.charAt(i));
        }
        System.out.println(reversed.toString());
    }
}

A dlorw

B world

C dlrow

D wrold

💡 Hint

The loop goes from the last character to the first.

The loop appends characters starting from the last index to the first, reversing the string.

🔧 Debug

advanced

Identify the error in string character access

What error will this Java code produce when run?

DSA Java

public class Main {
    public static void main(String[] args) {
        String s = "abc";
        char c = s.charAt(3);
        System.out.println(c);
    }
}

A NullPointerException

B StringIndexOutOfBoundsException

C No error, prints 'c'

D ArrayIndexOutOfBoundsException

💡 Hint

Check the valid index range for the string.

The string "abc" has indices 0,1,2. Accessing index 3 is out of range and throws StringIndexOutOfBoundsException.

❓ Predict Output

advanced

Output of nested loops with string character access

What is the output of this Java code?

DSA Java

public class Main {
    public static void main(String[] args) {
        String s = "abc";
        StringBuilder output = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            for (int j = 0; j <= i; j++) {
                output.append(s.charAt(j));
            }
            output.append(",");
        }
        System.out.println(output.toString());
    }
}

A a,ab,abc,

B a,ab,ac,

C a,aab,abc,

D a,b,abc,

💡 Hint

The inner loop prints characters from start up to current outer index.

For i=0 prints 'a', for i=1 prints 'ab', for i=2 prints 'abc', each followed by a comma.

❓ Predict Output

expert

Output after filtering vowels from string

What is the output of this Java code that filters vowels from a string?

DSA Java

public class Main {
    public static void main(String[] args) {
        String s = "education";
        StringBuilder vowels = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
                vowels.append(c);
            }
        }
        System.out.println(vowels.toString());
    }
}

A uioa

B education

C edctn

D euaio

💡 Hint

Only vowels are appended to the result string.

The code appends only vowels found in the string 'education' in order: e, u, a, i, o.

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