Check if Number is Prime in DSA Python

Mental Model

A prime number is a number greater than 1 that has no divisors other than 1 and itself. To check if a number is prime, we test if it can be divided evenly by any number from 2 up to its square root.

Analogy: Imagine you have a group of identical candies and want to split them evenly into smaller groups without leftovers. If you can only split them into one big group or single candies, then the number of candies is prime.

Number: 7
Check divisors: 2
7 % 2 != 0
No divisor found -> prime

Dry Run Walkthrough

Input: number = 7

Goal: Determine if 7 is a prime number

Step 1: Check if number is less than or equal to 1

7 > 1, continue checking

Why: Numbers less than or equal to 1 are not prime

Step 2: Check divisibility by 2

7 % 2 = 1 (not zero)

Why: If divisible, number is not prime; here it is not divisible

Step 3: Loop ends (limit = int(math.isqrt(7)) = 2)

No divisors found

Why: No divisor found up to sqrt(7), so number is prime

Result:

7 is prime

Annotated Code

DSA Python

import math

def is_prime(number: int) -> bool:
    if number <= 1:
        return False  # Numbers <= 1 are not prime
    limit = int(math.isqrt(number))  # Only check up to square root
    for divisor in range(2, limit + 1):
        if number % divisor == 0:
            return False  # Found a divisor, not prime
    return True  # No divisors found, number is prime

# Driver code
num = 7
if is_prime(num):
    print(f"{num} is prime")
else:
    print(f"{num} is not prime")

if number <= 1:

Check if number is too small to be prime

limit = int(math.isqrt(number))

Calculate the integer square root to limit divisor checks

for divisor in range(2, limit + 1):

Check divisors from 2 up to square root

if number % divisor == 0:

If divisible, number is not prime

return True

No divisors found, number is prime

Output

7 is prime

Complexity Analysis

Time: O(√n) because we only check divisors up to the square root of the number

Space: O(1) because we use a fixed amount of extra space regardless of input size

vs Alternative: Checking all numbers up to n would be O(n), which is slower; limiting to √n reduces checks significantly

Edge Cases

number = 1

Returns False because 1 is not prime

DSA Python

if number <= 1:

number = 2

Returns True because 2 is the smallest prime number

DSA Python

for divisor in range(2, limit + 1):

number = 0 or negative numbers

Returns False because primes are positive integers greater than 1

DSA Python

if number <= 1:

When to Use This Pattern

When you need to check if a number has no divisors other than 1 and itself, use the prime check pattern by testing divisibility up to the square root.

Common Mistakes

Mistake: Checking divisibility up to the number itself instead of square root
Fix: Limit divisor checks to the integer square root to improve efficiency

Mistake: Not handling numbers less than or equal to 1 correctly
Fix: Add a condition to return False for numbers <= 1

Summary

Checks if a number is prime by testing divisibility up to its square root.

Use when you need to quickly determine if a number has no divisors other than 1 and itself.

Only check divisors up to the square root because any larger divisor would have a smaller pair already checked.

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