Using scanf for input - Time & Space Complexity

Time Complexity: Using scanf for input

O(n)

Understanding Time Complexity

When we use scanf to get input in C, it takes some time to read the data. We want to understand how this reading time changes as we ask for more inputs.

How does the time to read inputs grow when we increase the number of inputs?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <stdio.h>

int main() {
    int n, x;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &x);
    }
    return 0;
}
    

This code reads an integer n and then reads n more integers from the user.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Reading an integer using scanf inside the loop.
• How many times: Exactly n times, where n is the number of inputs to read.

How Execution Grows With Input

Each time we increase n, we do one more input read. So the total reading time grows directly with n.

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

Pattern observation: The time grows in a straight line as we add more inputs.

Final Time Complexity

Time Complexity: O(n)

This means the time to read inputs grows directly in proportion to how many inputs we read.

Common Mistake

[X] Wrong: "Reading inputs with scanf always takes the same time no matter how many inputs."

[OK] Correct: Each input requires a separate read operation, so more inputs mean more time spent reading.

Interview Connect

Understanding how input reading scales helps you reason about program speed and efficiency, a useful skill in many coding challenges and real projects.

Self-Check

"What if we read inputs using a single line with multiple values instead of one by one? How would the time complexity change?"

Practice

1. What is the purpose of the scanf function in C?

easy

A. To read input from the user during program execution

B. To print output to the screen

C. To declare variables

D. To perform mathematical calculations

Solution

1. Step 1: Understand the role of scanf

   scanf is used to get input from the user while the program runs.

2. Step 2: Differentiate from other functions

   Functions like printf print output, but scanf reads input.

3. Final Answer:

   To read input from the user during program execution -> Option A

4. Quick Check:

   scanf reads input [OK]

Hint: scanf reads input, printf prints output [OK]

Common Mistakes:

• Confusing scanf with printf
• Thinking scanf declares variables
• Assuming scanf performs calculations

2. Which of the following is the correct way to read an integer variable num using scanf?

easy

A. scanf("%s", &num);

B. scanf("%d", num);

C. scanf("%f", &num);

D. scanf("%d", &num);

Solution

1. Step 1: Identify the correct format specifier for integer

   Use %d to read an integer.

2. Step 2: Use the address operator for variables

   Use &num to pass the address of the variable to scanf.

3. Final Answer:

   scanf("%d", &num); -> Option D

4. Quick Check:

   Integer input needs %d and &variable [OK]

Hint: Use & before variable except arrays [OK]

Common Mistakes:

• Omitting & before variable
• Using wrong format specifier like %f or %s
• Passing variable instead of its address

3. What will be the output of the following code if the user inputs 25?

int age;
scanf("%d", &age);
printf("Age: %d", age);

medium

A. Age: 0

B. Age: 25

C. Age: %d

D. Compilation error

Solution

1. Step 1: Understand input reading

   scanf("%d", &age); reads the integer 25 and stores it in age.

2. Step 2: Print the stored value

   printf("Age: %d", age); prints the string with the value of age, which is 25.

3. Final Answer:

   Age: 25 -> Option B

4. Quick Check:

   scanf stores input, printf prints it [OK]

Hint: scanf stores input, printf prints variable [OK]

Common Mistakes:

• Forgetting & in scanf causing runtime error
• Using wrong format specifier in printf
• Expecting scanf to print output

4. Identify the error in the following code snippet:

int x;
printf("Enter a number: ");
scanf("%d", x);
printf("You entered %d", x);

medium

A. Missing & before variable x in scanf

B. Wrong format specifier in scanf

C. Missing semicolon after printf

D. Variable x not declared

Solution

1. Step 1: Check scanf usage

   scanf("%d", x); is incorrect because it should use &x to pass the address.

2. Step 2: Confirm other parts are correct

   Format specifier is correct, semicolons are present, and x is declared.

3. Final Answer:

   Missing & before variable x in scanf -> Option A

4. Quick Check:

   scanf needs &variable for input [OK]

Hint: Always use & before variables in scanf except arrays [OK]

Common Mistakes:

• Omitting & causing segmentation fault
• Using wrong format specifier
• Assuming scanf prints output

5. You want to read a string (word) from the user into a character array name[20]. Which of the following scanf statements is correct and safe to avoid buffer overflow?

hard

A. scanf("%s", name);

B. scanf("%s", &name);

C. scanf("%19s", name);

D. scanf("%20s", &name);

Solution

1. Step 1: Understand string input with scanf

   To read a string into name, use %s without & because arrays decay to pointers.

2. Step 2: Prevent buffer overflow

   Limit input length to 19 characters using %19s to leave space for the null terminator.

3. Final Answer:

   scanf("%19s", name); -> Option C

4. Quick Check:

   Limit string input length with %19s [OK]

Hint: Use %19s for 20-char array to avoid overflow [OK]

Common Mistakes:

• Using & with arrays in scanf
• Not limiting input length causing overflow
• Using wrong format specifier for strings