If–else statement in C - Time & Space Complexity

Time Complexity: If-else statement

O(1)

Understanding Time Complexity

We want to see how the time to run an if-else statement changes as the input changes.

Does the program take longer if the input is bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int checkNumber(int n) {
    if (n > 0) {
        return 1;
    } else {
        return 0;
    }
}
    

This code checks if a number is positive and returns 1 or 0 accordingly.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single if-else check.
• How many times: Exactly once per function call.

How Execution Grows With Input

The time to run this code does not change with the size of the input number.

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

Pattern observation: The number of operations stays the same no matter how big the input is.

Final Time Complexity

Time Complexity: O(1)

This means the time to run the if-else statement stays constant, no matter the input size.

Common Mistake

[X] Wrong: "If the input number is bigger, the if-else will take longer to decide."

[OK] Correct: The if-else only checks the condition once, so the input size does not affect the time.

Interview Connect

Understanding that simple if-else checks run in constant time helps you explain how decisions in code affect performance.

Self-Check

"What if we added a loop inside the else block that runs n times? How would the time complexity change?"

Practice

1. What does an if-else statement do in C?

easy

A. It declares a variable.

B. It repeats a block of code multiple times.

C. It defines a new function.

D. It chooses between two actions based on a condition.

Solution

1. Step 1: Understand the purpose of if-else

   An if-else statement lets the program pick one of two paths depending on whether a condition is true or false.

2. Step 2: Match the description to the options

   Only It chooses between two actions based on a condition. describes choosing between two actions based on a condition, which is exactly what if-else does.

3. Final Answer:

   It chooses between two actions based on a condition. -> Option D

4. Quick Check:

   If-else = choose between two paths [OK]

Hint: If you want two choices, use if-else [OK]

Common Mistakes:

• Confusing if-else with loops
• Thinking if-else declares variables
• Mixing if-else with function definitions

2. Which of the following is the correct syntax for an if-else statement in C?

easy

A. if (condition) then { /* code */ } else { /* code */ }

B. if condition then { /* code */ } else { /* code */ }

C. if (condition) { /* code */ } else { /* code */ }

D. if condition { /* code */ } else { /* code */ }

Solution

1. Step 1: Recall C syntax rules for if-else

   In C, conditions must be inside parentheses and braces {} are used for code blocks.

2. Step 2: Check each option

   Only if (condition) { /* code */ } else { /* code */ } uses parentheses around the condition and braces correctly without extra keywords like 'then'.

3. Final Answer:

   if (condition) { /* code */ } else { /* code */ } -> Option C

4. Quick Check:

   Correct syntax uses parentheses and braces [OK]

Hint: Remember: if(condition) with braces {} [OK]

Common Mistakes:

• Omitting parentheses around condition
• Using 'then' keyword (not in C)
• Skipping braces for multiple statements

3. What will be the output of this C code?

#include 
int main() {
    int x = 10;
    if (x > 5) {
        printf("Greater\n");
    } else {
        printf("Smaller\n");
    }
    return 0;
}

medium

A. Greater

B. Smaller

C. 10

D. No output

Solution

1. Step 1: Evaluate the condition x > 5

   Variable x is 10, which is greater than 5, so the condition is true.

2. Step 2: Determine which block runs

   Since the condition is true, the code inside the if block runs, printing "Greater".

3. Final Answer:

   Greater -> Option A

4. Quick Check:

   10 > 5 is true, so prints "Greater" [OK]

Hint: Check if condition is true or false to pick output [OK]

Common Mistakes:

• Confusing which block runs on true condition
• Ignoring the newline character in printf
• Assuming else block runs when condition is true

4. Find the error in this if-else code snippet:

int a = 3;
if a > 0
    printf("Positive\n");
else
    printf("Non-positive\n");

medium

A. Missing semicolon after if statement

B. Missing parentheses around condition in if statement

C. Missing braces around printf statements

D. Incorrect else keyword

Solution

1. Step 1: Check syntax of if condition

   In C, the condition in an if statement must be inside parentheses (). Here, it is missing.

2. Step 2: Verify other parts

   Semicolons after printf are present, braces are optional for single statements, and else keyword is correct.

3. Final Answer:

   Missing parentheses around condition in if statement -> Option B

4. Quick Check:

   if condition needs parentheses ( ) [OK]

Hint: Always put condition inside parentheses ( ) [OK]

Common Mistakes:

• Forgetting parentheses in if condition
• Thinking braces are always required for single statements
• Misplacing else keyword

5. You want to write a program that prints "Even" if a number is even, and "Odd" if it is odd. Which if-else code correctly does this in C?

hard

A. if (num % 2 == 0) { printf("Even\n"); } else { printf("Odd\n"); }

B. if (num / 2 == 0) { printf("Even\n"); } else { printf("Odd\n"); }

C. if (num % 2 = 0) { printf("Even\n"); } else { printf("Odd\n"); }

D. if (num % 2 != 0) { printf("Even\n"); } else { printf("Odd\n"); }

Solution

1. Step 1: Understand how to check even numbers

   A number is even if dividing by 2 leaves remainder 0, which is checked by num % 2 == 0.

2. Step 2: Analyze each option

   if (num % 2 == 0) { printf("Even\n"); } else { printf("Odd\n"); } correctly uses modulus operator and equality check. if (num / 2 == 0) { printf("Even\n"); } else { printf("Odd\n"); } uses division instead of modulus. if (num % 2 = 0) { printf("Even\n"); } else { printf("Odd\n"); } uses assignment (=) instead of comparison (==). if (num % 2 != 0) { printf("Even\n"); } else { printf("Odd\n"); } reverses the logic.

3. Final Answer:

   if (num % 2 == 0) { printf("Even\n"); } else { printf("Odd\n"); } -> Option A

4. Quick Check:

   Use modulus == 0 to check even [OK]

Hint: Use num % 2 == 0 to test even numbers [OK]

Common Mistakes:

• Using division instead of modulus
• Using = instead of == in condition
• Reversing even and odd logic