Common wrapper methods in Java - Time & Space Complexity

Time Complexity: Common wrapper methods

O(1)

Understanding Time Complexity

We want to see how fast common wrapper methods run as the input size grows.

How does the time to run these methods change when we use bigger or smaller data?

Scenario Under Consideration

Analyze the time complexity of these wrapper method calls.

Integer num = Integer.valueOf(100);
String str = num.toString();
int val = num.intValue();
boolean isEqual = num.equals(100);

This code uses common wrapper methods to convert, compare, and get values.

Identify Repeating Operations

Look for loops or repeated work inside these methods.

• Primary operation: Simple method calls without loops.
• How many times: Each method runs once per call.

How Execution Grows With Input

These methods do a fixed amount of work regardless of input size.

Input Size (n)	Approx. Operations
10	About 1 operation
100	About 1 operation
1000	About 1 operation

Pattern observation: The time stays the same no matter how big the input is.

Final Time Complexity

Time Complexity: O(1)

This means the methods take the same short time no matter the input size.

Common Mistake

[X] Wrong: "Calling toString() or equals() takes longer if the number is bigger."

[OK] Correct: These methods do simple fixed steps, so size or value does not affect their speed.

Interview Connect

Knowing these methods run quickly helps you explain your code's efficiency clearly and confidently.

Self-Check

"What if we called toString() on a very large collection instead of a single number? How would the time complexity change?"

Practice

1. Which of the following methods converts a String to a primitive int in Java?

easy

A. Integer.valueOf(String s)

B. Integer.toString(int i)

C. Integer.parseInt(String s)

D. Integer.intValue()

Solution

1. Step 1: Understand method purpose

   parseInt converts a string directly to a primitive int.

2. Step 2: Differentiate from other methods

   valueOf returns an Integer object, not a primitive. toString converts int to string. intValue() is called on an Integer object to get primitive int.

3. Final Answer:

   Integer.parseInt(String s) -> Option C

4. Quick Check:

   parseInt returns int [OK]

Hint: parseInt returns primitive int, valueOf returns Integer object [OK]

Common Mistakes:

• Confusing parseInt and valueOf
• Using intValue() without an Integer object
• Thinking toString converts string to int

2. Which of the following is the correct syntax to convert an int variable num to an Integer object?

easy

A. Integer numObj = Integer.valueOf(num);

B. Integer numObj = Integer.parseInt(num);

C. Integer numObj = num.intValue();

D. Integer numObj = Integer.toString(num);

Solution

1. Step 1: Identify method to get Integer object

   valueOf converts primitive int to Integer object.

2. Step 2: Check other options

   parseInt returns primitive int, not Integer object. intValue() is a method of Integer, not int. toString returns String, not Integer.

3. Final Answer:

   Integer.valueOf(num) -> Option A

4. Quick Check:

   valueOf returns Integer object [OK]

Hint: Use valueOf to get Integer object from int [OK]

Common Mistakes:

• Using parseInt with int input
• Calling intValue() on primitive int
• Confusing toString with valueOf

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

String s = "123";
Integer obj = Integer.valueOf(s);
int n = obj.intValue();
System.out.println(n + 10);

medium

A. 133

B. Compilation error

C. NumberFormatException

D. 12310

Solution

1. Step 1: Convert String to Integer object

   Integer.valueOf("123") creates Integer object with value 123.

2. Step 2: Extract primitive int and add 10

   intValue() returns 123 as int. Adding 10 results in 133.

3. Final Answer:

   133 -> Option A

4. Quick Check:

   123 + 10 = 133 [OK]

Hint: valueOf + intValue() converts string to int for math [OK]

Common Mistakes:

• Thinking output is string concatenation
• Expecting NumberFormatException for valid number string
• Confusing intValue() usage

4. Identify the error in the following code snippet:

String s = "456";
int n = Integer.valueOf(s);
System.out.println(n);

medium

A. Missing semicolon after Integer.valueOf(s)

B. String s cannot be converted to Integer

C. Cannot assign Integer object to int directly

D. No error, code runs fine

Solution

1. Step 1: Check return type of Integer.valueOf

   valueOf returns an Integer object.

2. Step 2: Auto-unboxing allows assignment

   Java (5+) automatically unboxes Integer to int, so the code compiles and runs, printing 456.

3. Final Answer:

   No error, code runs fine -> Option D

4. Quick Check:

   valueOf(String) auto-unboxes to int [OK]

Hint: valueOf(String) auto-unboxes to int [OK]

Common Mistakes:

• Forgetting auto-unboxing feature
• Assuming valueOf returns primitive int
• Thinking String can't convert to Integer

5. You have a list of numeric strings: ["10", "20", "30", "40"]. Which code snippet correctly converts this list to a list of Integer objects using wrapper methods?

hard

A. List<Integer> nums = list.stream().map(String::parseInt).collect(Collectors.toList());

B. List<Integer> nums = list.stream().map(Integer::valueOf).collect(Collectors.toList());

C. List<Integer> nums = list.stream().map(s -> s.intValue()).collect(Collectors.toList());

D. List<Integer> nums = list.stream().map(s -> Integer.toString(s)).collect(Collectors.toList());

Solution

1. Step 1: Understand method return types

   Integer.valueOf(String) returns an Integer object, suitable for list of Integer.

2. Step 2: Check other options

   String::parseInt invalid - no parseInt on String. s.intValue() not a method on String. toString returns String, not Integer.

3. Final Answer:

   list.stream().map(Integer::valueOf).collect(Collectors.toList()) -> Option B

4. Quick Check:

   valueOf converts String to Integer object [OK]

Hint: Use valueOf to convert String to Integer object in streams [OK]

Common Mistakes:

• Thinking parseInt is a String method
• Calling intValue() on String
• Using toString instead of valueOf