String Basics and Memory Representation in DSA Java

Mental Model

A string is a sequence of characters stored one after another in memory, like beads on a string.

Analogy: Imagine a necklace where each bead is a letter. The necklace holds all letters in order, and you can read the message by looking at each bead one by one.

String: [H] -> [e] -> [l] -> [l] -> [o] -> null
Index:   0      1      2      3      4

Dry Run Walkthrough

Input: string: "Hello"

Goal: Understand how the string "Hello" is stored in memory as a sequence of characters

Step 1: Store character 'H' at index 0

[H] -> null

Why: Start storing the first character at the beginning of the string

Step 2: Store character 'e' at index 1

[H] -> [e] -> null

Why: Add the next character in sequence to maintain order

Step 3: Store character 'l' at index 2

[H] -> [e] -> [l] -> null

Why: Continue storing characters in order

Step 4: Store character 'l' at index 3

[H] -> [e] -> [l] -> [l] -> null

Why: Store repeated character maintaining sequence

Step 5: Store character 'o' at index 4

[H] -> [e] -> [l] -> [l] -> [o] -> null

Why: Complete storing all characters of the string

Result:

Final memory layout: [H] -> [e] -> [l] -> [l] -> [o] -> null

Annotated Code

DSA Java

public class StringMemory {
    public static void main(String[] args) {
        String s = "Hello";
        char[] chars = s.toCharArray();
        System.out.print("String characters in memory: ");
        for (int i = 0; i < chars.length; i++) {
            System.out.print("[" + chars[i] + "]");
            if (i < chars.length - 1) {
                System.out.print(" -> ");
            }
        }
        System.out.print(" -> null\n");
    }
}

char[] chars = s.toCharArray();

Convert string to array of characters to represent memory layout

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

Traverse each character in the array to print memory order

System.out.print("[" + chars[i] + "]");

Print current character in brackets to show individual storage

if (i < chars.length - 1) { System.out.print(" -> "); }

Print arrow to indicate next character in sequence

Output

String characters in memory: [H] -> [e] -> [l] -> [l] -> [o] -> null

Complexity Analysis

Time: O(n) because we process each character once to show memory layout

Space: O(n) because the character array stores all characters of the string

vs Alternative: Using a character array explicitly shows memory layout, while using string methods hides this detail

Edge Cases

Empty string ""

The character array is empty and prints no characters, just ' -> null'

DSA Java

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

String with one character "A"

Prints one character in brackets followed by ' -> null'

DSA Java

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

When to Use This Pattern

When you need to understand how strings are stored or manipulated at the character level, think of strings as arrays of characters in memory.

Common Mistakes

Mistake: Trying to modify a string directly as if it were an array
Fix: Convert the string to a character array first, modify the array, then create a new string if needed

Summary

Shows how a string is stored as a sequence of characters in memory.

Use when you want to understand or manipulate the individual characters of a string.

Remember a string is like a chain of characters stored one after another, not a single indivisible object.

Practice

1. What is the correct way to create a string literal in Java?

easy

A. String s = 'Hello';

B. String s = new String();

C. String s = "Hello";

D. String s = String();

Solution

1. Step 1: Understand string literal syntax

   In Java, string literals are enclosed in double quotes, so "Hello" is correct.

2. Step 2: Check each option

   String s = "Hello"; uses double quotes correctly. String s = new String(); creates an empty string object but not a literal. String s = 'Hello'; uses single quotes which are for characters, not strings. String s = String(); is invalid syntax.

3. Final Answer:

   String s = "Hello"; -> Option C

4. Quick Check:

   String literal uses double quotes = A [OK]

Hint: Use double quotes for string literals in Java [OK]

Common Mistakes:

• Using single quotes for strings
• Trying to call String() like a function
• Confusing new String() with literal

2. Which of the following is the correct syntax to compare two strings s1 and s2 for content equality in Java?

easy

A. if (s1.equals(s2))

B. if (s1 == s2)

C. if (s1 = s2)

D. if (s1.compare(s2))

Solution

1. Step 1: Understand string comparison in Java

   Using == compares references (memory addresses), not content. To compare content, use the equals() method.

2. Step 2: Evaluate each option

   if (s1 == s2) compares references, not content. if (s1.equals(s2)) correctly uses equals() method. if (s1 = s2) uses assignment (=) which is invalid in condition. if (s1.compare(s2)) uses a non-existent method compare().

3. Final Answer:

   if (s1.equals(s2)) -> Option A

4. Quick Check:

   Use equals() for string content comparison = B [OK]

Hint: Use equals() method to compare string contents [OK]

Common Mistakes:

• Using == to compare strings
• Using assignment operator in if condition
• Calling non-existent methods

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

String s1 = "hello";
String s2 = "hello";
String s3 = new String("hello");
System.out.println(s1 == s2);
System.out.println(s1 == s3);
System.out.println(s1.equals(s3));

medium

A. true
false
true

B. false
false
true

C. true
true
true

D. false
true
false

Solution

1. Step 1: Understand string pool and new String()

   s1 and s2 are string literals and point to the same object in the string pool, so s1 == s2 is true. s3 is a new String object, so s1 == s3 is false because they reference different objects.

2. Step 2: Check equals() method

   s1.equals(s3) compares content and both have "hello", so it returns true.

3. Final Answer:

   true false true -> Option A

4. Quick Check:

   == compares references, equals() compares content = C [OK]

Hint: == checks memory, equals() checks string content [OK]

Common Mistakes:

• Assuming new String() shares reference
• Confusing == with equals()
• Ignoring string pool behavior

4. Identify the error in the following code snippet:

String s = "test";
if (s == "test") {
    System.out.println("Match");
} else {
    System.out.println("No Match");
}

medium

A. System.out.println syntax is incorrect

B. Missing semicolon after string declaration

C. String literals cannot be compared

D. Using == to compare strings can cause unexpected results

Solution

1. Step 1: Check string comparison method

   The code uses == to compare strings, which compares references, not content. This can cause unexpected results if strings are not from the pool.

2. Step 2: Verify other syntax

   There is no missing semicolon, string literals can be compared with equals(), and System.out.println syntax is correct.

3. Final Answer:

   Using == to compare strings can cause unexpected results -> Option D

4. Quick Check:

   Use equals() for string comparison, not == = A [OK]

Hint: Avoid == for strings; use equals() instead [OK]

Common Mistakes:

• Thinking == compares string content
• Ignoring string pool behavior
• Assuming syntax errors where none exist

5. Given the following code, what will be the output?

String a = "Java";
String b = "Ja" + "va";
String c = new String("Java");
System.out.println(a == b);
System.out.println(a == c);
System.out.println(a.equals(c));

Explain why the outputs are as they are.

hard

A. false
true
false

B. true
false
true

C. true
true
true

D. false
false
true

Solution

1. Step 1: Analyze string concatenation with literals

   Strings "Ja" and "va" are literals concatenated at compile time, so b points to the same string pool object as a. Therefore, a == b is true.

2. Step 2: Understand new String() behavior

   c is a new String object created at runtime, so a == c is false because they reference different objects. However, a.equals(c) compares content and returns true.

3. Final Answer:

   true false true -> Option B

4. Quick Check:

   Compile-time concat shares pool, new String() creates new object = D [OK]

Hint: Compile-time concat shares pool; new String() creates new object [OK]

Common Mistakes:

• Assuming runtime concat shares pool
• Confusing == with equals()
• Ignoring compile-time optimization