String Traversal and Character Access in DSA Java

Mental Model

A string is a sequence of characters you can look at one by one from start to end.

Analogy: Imagine reading a sentence letter by letter from left to right to understand or find something.

"H" -> "e" -> "l" -> "l" -> "o" -> null
 ↑

Dry Run Walkthrough

Input: string: "Hello", print each character in order

Goal: Print each character one by one from the string

Step 1: Access character at index 0

"H" -> "e" -> "l" -> "l" -> "o" -> null
 ↑

Why: Start from the first character to begin traversal

Step 2: Print character 'H'

"H" -> "e" -> "l" -> "l" -> "o" -> null
   ↑

Why: Output the current character before moving on

Step 3: Move to index 1 and access character

"H" -> "e" -> "l" -> "l" -> "o" -> null
     ↑

Why: Advance to next character to continue traversal

Step 4: Print character 'e'

"H" -> "e" -> "l" -> "l" -> "o" -> null
       ↑

Why: Output the current character

Step 5: Repeat for indices 2, 3, 4 printing 'l', 'l', 'o'

"H" -> "e" -> "l" -> "l" -> "o" -> null
             ↑

Why: Continue until all characters are printed

Result:

Printed characters: H
e
l
l
o
Final string traversal complete

Annotated Code

DSA Java

public class StringTraversal {
    public static void traverseString(String input) {
        for (int i = 0; i < input.length(); i++) {
            char currentChar = input.charAt(i); // access character at index i
            System.out.println(currentChar); // print current character
        }
        System.out.println("Final string traversal complete");
    }

    public static void main(String[] args) {
        String testString = "Hello";
        traverseString(testString);
    }
}

char currentChar = input.charAt(i); // access character at index i

access character at current index to process it

System.out.println(currentChar); // print current character

output the character to show traversal progress

for (int i = 0; i < input.length(); i++) {

loop through each character index from start to end

Output

H

e

l

l

o

Final string traversal complete

Complexity Analysis

Time: O(n) because we visit each character once where n is string length

Space: O(1) because we only store one character at a time during traversal

vs Alternative: Compared to copying the string or using extra data structures, this direct access is more efficient in space and time

Edge Cases

empty string ""

loop does not run, no characters printed, just final message

DSA Java

for (int i = 0; i < input.length(); i++) {

string with one character "A"

prints single character then final message

DSA Java

for (int i = 0; i < input.length(); i++) {

When to Use This Pattern

When you need to process or examine each character in a string one by one, use string traversal with character access.

Common Mistakes

Mistake: Using index out of bounds by looping beyond string length
Fix: Ensure loop condition uses input.length() as upper bound, not greater

Mistake: Trying to modify characters directly in string (strings are immutable)
Fix: Use a new string or character array if modification is needed

Summary

It goes through each character in a string one by one.

Use it when you want to read or process every letter in a word or sentence.

Remember strings are like a row of boxes you open one at a time from left to right.

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