String Traversal and Character Access in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does it mean to traverse a string in programming?

Traversing a string means going through each character one by one, usually from the first character to the last, to read or process them.

beginner

How do you access the character at a specific position in a string in Java?

You use the method charAt(index), where index is the position starting from 0 for the first character.

beginner

What is the index of the first character in a Java string?

The first character of a Java string is at index 0 because Java uses zero-based indexing.

intermediate

Why is zero-based indexing used in string traversal?

Zero-based indexing starts counting from 0, which matches how memory addresses work in computers and makes calculations simpler.

intermediate

What happens if you try to access a character at an index outside the string length in Java?

Java throws a StringIndexOutOfBoundsException error because the index is invalid and does not point to any character.

In Java, what does str.charAt(3) return?

A An error because index 3 is invalid

B The third character of the string <code>str</code>

C The first character of the string <code>str</code>

D The fourth character of the string <code>str</code>

Java strings use zero-based indexing, so index 3 is the fourth character.

What is the index of the last character in a string of length 10?

A 9

B 10

C 0

D 11

The last character is at index length - 1, so 10 - 1 = 9.

Which method is used to find the length of a string in Java?

A length

B size()

C length()

D count()

Java strings use the length() method to get the number of characters.

What will happen if you try to access str.charAt(-1) in Java?

A Returns the first character

B Throws StringIndexOutOfBoundsException

C Returns null

D Returns the last character

Negative indexes are invalid and cause an exception in Java.

Which loop is commonly used to traverse all characters in a string?

A for loop with index from 0 to length-1

B while loop with index starting at length

C do-while loop starting at length

D for loop with index from 1 to length

A for loop from 0 to length-1 covers all characters in order.

Explain how to traverse a string and access each character in Java.

Think about how to visit every character one by one.

What errors can occur when accessing characters in a string and how to avoid them?

Consider what happens if you ask for a character outside the string.

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