Assignment operators in C - 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 assignments changes as the program runs with bigger inputs.

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 with addition
}
return sum;
    

This code adds numbers from 0 up to n-1 and stores the result in sum.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The assignment 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 gets bigger, the number of assignments grows directly with n.

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

Pattern observation: The number of assignments grows in a straight line as n increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the assignments grows directly with the size of the input.

Common Mistake

[X] Wrong: "Assignment operations inside a loop don't affect time much because they are simple."

[OK] Correct: Even simple assignments add up when repeated many times, so they do affect total time.

Interview Connect

Understanding how assignments inside loops affect time helps you explain code efficiency clearly and confidently.

Self-Check

"What if we replaced the assignment sum += i with a function call inside the loop? How would the time complexity change?"

Practice

1. Which of the following assignment operators adds the right operand to the left operand and then assigns the result to the left operand?

easy

A. -=

B. +=

C. *=

D. /=

Solution

1. Step 1: Understand the meaning of += operator

   The operator += adds the value on the right to the variable on the left and stores the result back in the left variable.

2. Step 2: Compare with other operators

   Other operators like -= subtract, *= multiply, and /= divide, so they do not add.

3. Final Answer:

   += -> Option B

4. Quick Check:

   += means add and assign [OK]

Hint: Remember '+=' means add then assign to left variable [OK]

Common Mistakes:

• Confusing '+=' with '=' which just assigns
• Mixing '+=' with '-=' or '*='
• Thinking '+=' only adds without assignment

2. Which of the following is the correct syntax using the compound assignment operator to multiply variable x by 5 and assign the result back to x?

easy

A. x = x *5;

B. x =* 5;

C. x = *5;

D. x *= 5;

Solution

1. Step 1: Identify correct compound assignment syntax

   The correct syntax for multiplying and assigning is 'x *= 5;'. It combines multiplication and assignment.

2. Step 2: Check other options for syntax errors

   x =* 5; 'x =* 5;' is invalid syntax. x = *5; 'x = *5;' is invalid because '*5' is not a valid expression. x = x *5; 'x = x *5;' is correct but not a compound assignment operator.

3. Final Answer:

   x *= 5; -> Option D

4. Quick Check:

   '*=' is multiply and assign operator [OK]

Hint: Use '*=' to multiply and assign in one step [OK]

Common Mistakes:

• Writing '=*' instead of '*='
• Forgetting space or semicolon
• Using 'x = *5;' which is invalid

3. What is the output of the following code?

int a = 10;
a += 5;
a -= 3;
printf("%d", a);

medium

A. 8

B. 15

C. 12

D. 10

Solution

1. Step 1: Calculate after first operation 'a += 5;'

   Starting with a = 10, adding 5 gives a = 15.

2. Step 2: Calculate after second operation 'a -= 3;'

   Subtracting 3 from 15 gives a = 12.

3. Final Answer:

   12 -> Option C

4. Quick Check:

   10 + 5 - 3 = 12 [OK]

Hint: Apply each operator step-by-step to track value [OK]

Common Mistakes:

• Adding and subtracting in wrong order
• Confusing '+=' and '-=' effects
• Ignoring initial value of variable

4. Identify the error in the following code snippet:

int x = 7;
x +== 3;
printf("%d", x);

medium

A. Syntax error due to '+==' operator

B. Runtime error due to invalid operation

C. No error, output is 10

D. Logical error, output is wrong

Solution

1. Step 1: Check the operator '+==' in the code

   The operator '+==' does not exist in C. The correct operator for addition assignment is '+='.

2. Step 2: Understand the effect of invalid operator

   Using '+==' causes a syntax error and the code will not compile.

3. Final Answer:

   Syntax error due to '+==' operator -> Option A

4. Quick Check:

   Invalid operator '+==' causes syntax error [OK]

Hint: Use '+=' not '+==' for addition assignment [OK]

Common Mistakes:

• Typing '+==' instead of '+='
• Assuming '+==' works like '=='
• Ignoring compiler error messages

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

int result = 20;
result /= 4;
result *= 3;
result %= 5;
result += 2;
printf("%d", result);

hard

A. 2

B. 3

C. 4

D. 1

Solution

1. Step 1: Perform division assignment 'result /= 4;'

   Starting with result = 20, dividing by 4 gives result = 5.

2. Step 2: Perform multiplication assignment 'result *= 3;'

   Multiplying 5 by 3 gives result = 15.

3. Step 3: Perform modulus assignment 'result %= 5;'

   15 modulo 5 is 0, so result = 0.

4. Step 4: Perform addition assignment 'result += 2;'

   Adding 2 to 0 gives result = 2.

5. Final Answer:

   2 -> Option A

6. Quick Check:

   20/4=5, 5*3=15, 15%5=0, 0+2=2 [OK]

Hint: Apply each operator in order carefully [OK]

Common Mistakes:

• Skipping modulus step or calculating it wrong
• Mixing up order of operations
• Forgetting that %= gives remainder