Check if Number is Prime in DSA Python - Time & Space Complexity

Time Complexity: Check if Number is Prime

O(n)

Understanding Time Complexity

We want to understand how the time to check if a number is prime grows as the number gets bigger.

Specifically, how many steps does it take to decide if a number has any divisors?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def is_prime(n: int) -> bool:
    if n <= 1:
        return False
    for i in range(2, n):
        if n % i == 0:
            return False
    return True
    

This code checks if a number n is prime by testing all numbers from 2 up to n-1 for divisibility.

Identify Repeating Operations

• Primary operation: The for-loop that tests divisibility from 2 to n-1.
• How many times: Up to n-2 times in the worst case (when n is prime).

How Execution Grows With Input

As the number n grows, the number of divisibility checks grows roughly the same as n.

Input Size (n)	Approx. Operations
10	Up to 8 checks
100	Up to 98 checks
1000	Up to 998 checks

Pattern observation: The number of steps grows roughly in a straight line with n.

Final Time Complexity

Time Complexity: O(n)

This means the time to check if a number is prime grows directly with the size of the number.

Common Mistake

[X] Wrong: "Checking divisibility up to n/2 is enough and faster than checking up to n."

[OK] Correct: While checking up to n/2 reduces steps, it is still linear and not optimal. The best known simple method checks only up to the square root of n, which is much faster.

Interview Connect

Understanding this time complexity helps you explain how to improve prime checks and shows you can analyze simple loops clearly.

Self-Check

"What if we change the loop to run only up to the square root of n? How would the time complexity change?"

Practice

1. What does it mean for a number to be prime?

easy

A. It has exactly two distinct positive divisors: 1 and itself.

B. It is divisible by 2 only.

C. It is an even number greater than 2.

D. It has more than two divisors.

Solution

1. Step 1: Understand the definition of prime number

   A prime number is a number greater than 1 that has no divisors other than 1 and itself.

2. Step 2: Compare options with definition

   Only It has exactly two distinct positive divisors: 1 and itself. correctly states this definition.

3. Final Answer:

   It has exactly two distinct positive divisors: 1 and itself. -> Option A

4. Quick Check:

   Prime number = two divisors [OK]

Hint: Prime means only divisible by 1 and itself [OK]

Common Mistakes:

• Thinking even numbers are always prime
• Confusing prime with odd numbers
• Believing prime numbers have multiple divisors

2. Which of the following Python code snippets correctly checks if a number n is prime?

easy

A. for i in range(2, n): if n % i != 0: return False; return True

B. for i in range(1, n): if n % i == 0: return False; return True

C. for i in range(2, n): if n % i == 0: return False; return True

D. for i in range(2, n): if n % i == 1: return False; return True

Solution

1. Step 1: Analyze the loop range and condition

   for i in range(2, n): if n % i == 0: return False; return True checks divisibility from 2 to n-1, which is correct for prime checking.

2. Step 2: Check the condition for returning False

   If any divisor divides n evenly (remainder 0), return False; else True after loop ends.

3. Final Answer:

   for i in range(2, n): if n % i == 0: return False; return True -> Option C

4. Quick Check:

   Check divisors from 2 to n-1 for remainder 0 [OK]

Hint: Check divisors from 2 to n-1 for zero remainder [OK]

Common Mistakes:

• Starting loop from 1 instead of 2
• Checking for remainder not equal to zero
• Checking if remainder equals 1 instead of 0

3. What is the output of the following code when n = 11?

def is_prime(n):
    if n <= 1:
        return False
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return False
    return True

print(is_prime(11))

medium

A. True

B. False

C. None

D. Error

Solution

1. Step 1: Check if n is less than or equal to 1

   Since 11 > 1, skip the first return False.

2. Step 2: Loop from 2 to sqrt(11) + 1 (which is 4)

   Check divisibility by 2 and 3; 11 % 2 != 0 and 11 % 3 != 0, so loop ends without returning False.

3. Step 3: Return True after loop

   No divisor found, so function returns True.

4. Final Answer:

   True -> Option A

5. Quick Check:

   11 is prime, so output is True [OK]

Hint: Check divisors up to sqrt(n) for efficiency [OK]

Common Mistakes:

• Looping up to n instead of sqrt(n)
• Returning False too early
• Not handling n <= 1 case

4. Find the error in this prime checking function and select the correct fix:

def is_prime(n):
    if n < 2:
        return False
    for i in range(2, n):
        if n % i == 0:
            return False
        else:
            return True

medium

A. Change range to start from 1 instead of 2

B. Change if condition to n % i != 0

C. Add return True inside the else block

D. Remove the else block inside the loop to return True after loop ends

Solution

1. Step 1: Identify the problem with else inside the loop

   The else causes the function to return True on first non-divisor, which is incorrect.

2. Step 2: Fix by removing else and returning True after loop

   Return True only after checking all divisors without finding any.

3. Final Answer:

   Remove the else block inside the loop to return True after loop ends -> Option D

4. Quick Check:

   Return True only after full divisor check [OK]

Hint: Return True after loop, not inside it [OK]

Common Mistakes:

• Returning True too early inside loop
• Starting loop from 1
• Changing condition to wrong remainder check

5. You want to check if a large number n is prime efficiently. Which approach is best?

hard

A. Check divisibility from 2 to n-1

B. Check divisibility from 2 to sqrt(n)

C. Check divisibility only by 2 and 3

D. Check divisibility from 1 to n

Solution

1. Step 1: Understand the efficiency of divisor checking

   Checking up to n-1 or n is slow for large n; checking only 2 and 3 misses other divisors.

2. Step 2: Use the mathematical fact that divisors come in pairs

   If n has a divisor greater than sqrt(n), it must have a corresponding divisor less than sqrt(n).

3. Step 3: Check divisibility only up to sqrt(n)

   This reduces the number of checks drastically and is the best approach.

4. Final Answer:

   Check divisibility from 2 to sqrt(n) -> Option B

5. Quick Check:

   Divisors up to sqrt(n) suffice for prime check [OK]

Hint: Check divisors only up to square root of n [OK]

Common Mistakes:

• Checking all numbers up to n
• Checking only 2 and 3
• Starting divisor check from 1