Assignment operators in Java - Time & Space Complexity

Time Complexity: Assignment operators

O(n)

Understanding Time Complexity

Let's see how assignment operators affect the time it takes for a program to run.

We want to know how the number of steps changes when using assignments in code.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int sum = 0;
for (int i = 0; i < n; i++) {
    sum += i;  // assignment operator +=
}
System.out.println(sum);
    

This code adds numbers from 0 up to n-1 and stores the result using an assignment operator.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The assignment operation sum += i inside the loop.
• How many times: It runs once for each number from 0 to n-1, so n times.

How Execution Grows With Input

As n grows, the number of assignments grows in the same way.

Input Size (n)	Approx. Operations
10	10 assignments
100	100 assignments
1000	1000 assignments

Pattern observation: The number of assignments grows directly with n, doubling n doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the assignments grows in a straight line with the input size.

Common Mistake

[X] Wrong: "Assignment operators make the code run faster or slower in a way that changes the overall time complexity."

[OK] Correct: Assignment operators themselves are simple steps. The total time depends on how many times they run, not the operator type.

Interview Connect

Understanding how simple operations like assignments scale helps you explain code efficiency clearly and confidently.

Self-Check

"What if we replaced the loop with nested loops using assignment operators? How would the time complexity change?"

Practice

1. What does the assignment operator += do in Java?

easy

A. Divides the left variable by the right value

B. Subtracts the right value from the left variable

C. Multiplies the left variable by the right value

D. Adds the right value to the left variable and stores the result in the left variable

Solution

1. Step 1: Understand the += operator

   The += operator adds the right-hand side value to the left-hand side variable.

2. Step 2: Store the result back in the left variable

   The sum replaces the original value in the left variable.

3. Final Answer:

   Adds the right value to the left variable and stores the result in the left variable -> Option D

4. Quick Check:

   x += y means x = x + y [OK]

Hint: Think: add then assign back to the same variable [OK]

Common Mistakes:

• Confusing += with subtraction or multiplication
• Thinking it only adds without assignment
• Mixing up the order of operands

2. Which of the following is the correct syntax to multiply a variable num by 5 and assign the result back to num?

easy

A. num = * 5;

B. num *= 5;

C. num = num *;

D. num = 5 *;

Solution

1. Step 1: Identify the correct compound assignment syntax

   The operator *= multiplies and assigns in one step.

2. Step 2: Check the variable and value placement

   The correct form is num *= 5; which means num = num * 5;.

3. Final Answer:

   num *= 5; -> Option B

4. Quick Check:

   Use *= for multiply and assign [OK]

Hint: Look for operator before equals for combined action [OK]

Common Mistakes:

• Placing * after = without variable
• Leaving out the variable on right side
• Using invalid syntax like num = * 5

3. What is the output of this Java code?

int a = 10;
a -= 3;
a += 2;
System.out.println(a);

medium

A. 9

B. 11

C. 5

D. 15

Solution

1. Step 1: Calculate a -= 3;

   Subtract 3 from 10: 10 - 3 = 7, so a becomes 7.

2. Step 2: Calculate a += 2;

   Add 2 to 7: 7 + 2 = 9, so a becomes 9.

3. Step 3: Check the printed value

   The code prints the current value of a, which is 9.

4. Final Answer:

   9 -> Option A

5. Quick Check:

   10 - 3 + 2 = 9 [OK]

Hint: Perform each assignment step in order carefully [OK]

Common Mistakes:

• Adding before subtracting
• Mixing up += and -= effects
• Printing before assignment changes

4. Find the error in this code snippet:

int x = 5;
x + = 3;
System.out.println(x);

medium

A. Variable x is not initialized

B. Missing semicolon after System.out.println(x)

C. The operator + = has a space and is invalid

D. The code will print 8 without errors

Solution

1. Step 1: Check the operator syntax

   The operator += must not have spaces; + = is invalid.

2. Step 2: Confirm other syntax elements

   Variable x is initialized and semicolons are correct.

3. Final Answer:

   The operator + = has a space and is invalid -> Option C

4. Quick Check:

   No spaces allowed in compound assignment operators [OK]

Hint: Compound operators have no spaces, write as one token [OK]

Common Mistakes:

• Adding spaces inside operators
• Assuming code runs despite syntax errors
• Confusing initialization errors with operator errors

5. Given the code below, what is the final value of total?

int total = 20;
int discount = 5;
total -= discount *= 2;
System.out.println(total);

hard

A. 10

B. 30

C. 15

D. 5

Solution

1. Step 1: Evaluate discount *= 2;

   This multiplies discount by 2: 5 * 2 = 10, so discount becomes 10.

2. Step 2: Calculate total -= discount;

   Subtract discount (now 10) from total (20): 20 - 10 = 10.

3. Step 3: Confirm final value

   The final value of total is 10, which is printed.

4. Final Answer:

   10 -> Option A

5. Quick Check:

   discount = 5*2=10; total = 20-10=10 [OK]

Hint: Evaluate right side first in compound assignments [OK]

Common Mistakes:

• Ignoring that discount changes before subtraction
• Subtracting original discount value (5) instead of updated (10)
• Confusing order of operations in assignment