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

Time Complexity: Check if Number is Prime

O(√n)

Understanding Time Complexity

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

How does the number size affect the steps we take to decide if it is prime?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn is_prime(n: u32) -> bool {
    if n <= 1 {
        return false;
    }
    for i in 2..=((n as f64).sqrt() as u32) {
        if n % i == 0 {
            return false;
        }
    }
    true
}
    

This code checks if a number is prime by testing divisibility from 2 up to the square root of the number.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop checking divisibility from 2 to √n.
• How many times: Approximately √n times in the worst case.

How Execution Grows With Input

As the number n grows, the number of checks grows roughly with the square root of n.

Input Size (n)	Approx. Operations
10	About 3 checks (2,3)
100	About 10 checks (2 to 10)
1000	About 31 checks (2 to 31)

Pattern observation: The number of steps grows slower than n, closer to the square root of n.

Final Time Complexity

Time Complexity: O(√n)

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

Common Mistake

[X] Wrong: "We must check all numbers from 2 up to n-1 to find if n is prime."

[OK] Correct: Checking up to n-1 is unnecessary because if n has a divisor larger than its square root, it must also have a smaller one. So checking up to √n is enough and faster.

Interview Connect

Understanding this time complexity helps you explain efficient prime checking, a common building block in many problems. It shows you can find smarter ways to reduce work, a key skill in coding interviews.

Self-Check

"What if we checked divisibility only by prime numbers up to √n? How would the time complexity change?"

Practice

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

easy

A. It has more than two divisors.

B. It is divisible by 2 only.

C. It is an even number greater than 2.

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

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 D

4. Quick Check:

   Prime = two divisors [OK]

Hint: Prime numbers have only 1 and itself as divisors [OK]

Common Mistakes:

• Thinking even numbers are always prime
• Confusing prime with odd numbers
• Assuming 1 is prime

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

easy

A. for i in 1..n { if n % i == 0 { return false; } } true

B. for i in 2..n { if n % i == 0 { return false; } } true

C. for i in 2..=n { if n % i == 0 { return false; } } true

D. for i in 2..n { if n % i != 0 { return false; } } true

Solution

1. Step 1: Analyze the loop range for divisor checking

   We check divisors from 2 up to n-1 (exclusive of n) to find if any divides n.

2. Step 2: Check each option's loop and condition

   for i in 2..n { if n % i == 0 { return false; } } true correctly loops from 2 to n-1 and returns false if divisor found; others have errors like including n or wrong condition.

3. Final Answer:

   for i in 2..n { if n % i == 0 { return false; } } true -> Option B

4. Quick Check:

   Loop 2 to n-1 and check divisibility [OK]

Hint: Loop from 2 to n-1 to check divisors [OK]

Common Mistakes:

• Including n in the loop range
• Starting loop from 1 instead of 2
• Using wrong condition for divisibility

3. What is the output of this Rust function call: is_prime(11)?

fn is_prime(n: u32) -> bool {
    if n <= 1 { return false; }
    for i in 2..n {
        if n % i == 0 {
            return false;
        }
    }
    true
}

medium

A. true

B. false

C. panic at runtime

D. compile error

Solution

1. Step 1: Check input and loop behavior for n=11

   11 is greater than 1, so the function proceeds to loop from 2 to 10 checking divisibility.

2. Step 2: Test divisibility of 11 by numbers 2 to 10

   11 is not divisible by any number between 2 and 10, so loop completes without returning false.

3. Final Answer:

   true -> Option A

4. Quick Check:

   11 is prime, so function returns true [OK]

Hint: Prime numbers return true after no divisors found [OK]

Common Mistakes:

• Assuming 11 is not prime
• Confusing loop range
• Expecting runtime errors incorrectly

4. Identify the bug in this Rust function that checks if a number is prime:

fn is_prime(n: u32) -> bool {
    if n < 2 { return false; }
    for i in 2..=n {
        if n % i == 0 {
            return false;
        }
    }
    true
}

medium

A. Missing return type in function signature

B. Condition should be n <= 1 instead of n < 2

C. Loop includes n, causing always false for n > 1

D. Loop should start from 1 instead of 2

Solution

1. Step 1: Analyze the loop range 2..=n

   The loop includes n itself, so n % n == 0 is always true, causing immediate false return.

2. Step 2: Understand correct loop range

   Loop should be 2..n (exclusive of n) to avoid checking n against itself.

3. Final Answer:

   Loop includes n, causing always false for n > 1 -> Option C

4. Quick Check:

   Loop must exclude n to avoid false negatives [OK]

Hint: Exclude n from divisor loop to avoid false negatives [OK]

Common Mistakes:

• Including n in divisor checks
• Wrong loop start index
• Misunderstanding return conditions

5. You want to optimize prime checking for large numbers by only checking divisors up to the square root of n. Which Rust code snippet correctly implements this?

hard

A. for i in 2..=((n as f64).sqrt() as u32) { if n % i == 0 { return false; } } true

B. for i in 2..n { if n % i == 0 { return false; } } true

C. for i in 2..=(n / 2) { if n % i == 0 { return false; } } true

D. for i in 1..=((n as f64).sqrt() as u32) { if n % i == 0 { return false; } } true

Solution

1. Step 1: Understand optimization using square root

   Checking divisors up to sqrt(n) is enough because larger factors pair with smaller ones.

2. Step 2: Analyze each option's loop range and correctness

   for i in 2..=((n as f64).sqrt() as u32) { if n % i == 0 { return false; } } true correctly casts n to float, computes sqrt, casts back to u32, and loops up to that value.

3. Final Answer:

   for i in 2..=((n as f64).sqrt() as u32) { if n % i == 0 { return false; } } true -> Option A

4. Quick Check:

   Loop up to sqrt(n) for efficient prime check [OK]

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

Common Mistakes:

• Checking divisors beyond sqrt(n)
• Starting loop from 1 instead of 2
• Incorrect casting causing errors