Prime Number Checker - Problem

Given a positive integer n, determine whether it is a prime number.

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

Examples:

2, 3, 5, 7, 11 are prime numbers
4, 6, 8, 9, 10 are not prime numbers

Optimize your solution by checking only up to the square root of n.

Input & Output

Example 1 — Small Prime Number

$ Input: n = 17

› Output: true

💡 Note: 17 is a prime number. We only need to check divisors from 2 to √17 ≈ 4. None of 2, 3, or 4 divide 17 evenly, so it's prime.

Example 2 — Composite Number

$ Input: n = 15

› Output: false

💡 Note: 15 is not prime because it can be divided by 3 (15 ÷ 3 = 5) and 5 (15 ÷ 5 = 3). We find the divisor 3 when checking up to √15 ≈ 3.87.

Example 3 — Edge Case Small Number

$ Input: n = 2

› Output: true

💡 Note: 2 is the smallest prime number. It's only divisible by 1 and itself.

Constraints

1 ≤ n ≤ 10⁹

Visualization

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Asked in

G Google 12 M Microsoft 8 a Amazon 15

The key insight is that if a number n has divisors, at least one must be ≤ √n. Best approach is the square root optimization with Time: O(√n), Space: O(1)

Common Approaches

✓ Brute Force Division

⏱️ Time: O(n) Space: O(1)

Test every number from 2 to n-1 to see if any divides n evenly. If we find any divisor, the number is not prime. If we check all numbers without finding a divisor, it's prime.

Square Root Optimization

⏱️ Time: O(√n) Space: O(1)

Instead of checking all numbers up to n-1, we only check up to √n. This works because if n has a divisor greater than √n, it must also have a corresponding divisor less than √n.

Brute Force Division — Algorithm Steps

Handle edge cases: numbers ≤ 1 are not prime
Loop through all numbers from 2 to n-1
If any number divides n evenly, return false
If no divisors found, return true

Visualization

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Step-by-Step Walkthrough

Start Check

Begin testing from divisor 2

Test All

Check every number up to n-1

Result

Prime if no divisors found

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>

bool solution(int n) {
    if (n <= 1) {
        return false;
    }
    
    for (int i = 2; i < n; i++) {
        if (n % i == 0) {
            return false;
        }
    }
    
    return true;
}

int main() {
    int n;
    scanf("%d", &n);
    
    bool result = solution(n);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We check every number from 2 to n-1 in the worst case

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

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Prime Number Checker - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Division — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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