A prime number is a natural number greater than 1 and is completely divisible only by 1 and itself. In this article, we will learn how to check whether the given number is a prime number or not in C.
Examples:
Input: n = 29
Output: 29 is Prime
Explanation: 29 has no divisors other than 1 and 29 itself. Hence, it is a prime number.Input: n = 15
Output: 15 is NOT prime
Explanation: 15 has divisors other than 1 and 15 (i.e., 3 and 5). Hence, it is not a prime number.
We can check whether a number is prime using various approaches:
Table of Content
Brute Force Method - O(n) Time
We can check whether the number is prime or not by iterating in the range from 1 to n using loops. We will count the number of divisors. If there are more than 2 divisor (including 1 and n) then the given number n is not prime, else n is prime. This method is known as trial division method.
Example:
#include <stdbool.h>
#include <stdio.h>
int main() {
int n = 29;
int cnt = 0;
// If number is less than/equal to 1,
// it is not prime
if (n <= 1)
printf("%d is NOT prime", n);
else {
// Count the all divisors of
// given number
for (int i = 1; i <= n; i++) {
// Check n is divided by
// i or not
if (n % i == 0)
cnt++;
}
// If n is divisible by more than 2 numbers
// then it is not prime
if (cnt > 2)
printf("%d is NOT prime", n);
// else it is prime
else
printf("%d is prime", n);
}
return 0;
}
Output
29 is prime
The time complexity of the above program is O(n) because we iterate from 1 to n.
Optimized Approach - O(√n) Time
To optimize the above approach, we use a mathematical property which states that,
The smallest factor of a number greater than one cannot be greater than the square root of that number.
Using this, we can reduce the numbers to be checked from N to √N making it much more efficient than above approach.
Example:
#include <math.h>
#include <stdbool.h>
#include <stdio.h>
int main() {
int n = 29;
int cnt = 0;
// If number is less than/equal to 1,
// it is not prime
if (n <= 1)
printf("%d is NOT prime", n);
else {
// Check how many numbers divide n in
// range 2 to sqrt(n)
for (int i = 2; i * i <= n; i++) {
if (n % i == 0)
cnt++;
}
// if cnt is greater than 0 then n is
// not prime
if (cnt > 0)
printf("%d is NOT prime", n);
// else n is prime
else
printf("%d is prime", n);
}
return 0;
}
Output
29 is prime
The time complexity of the above program is O(√n) because we iterate only √n times.
We can further optimize the above approach by skipping all even numbers greater than 2. Since the only even prime number is 2, we can skip all even numbers between 3 and √n.
Example:
#include <math.h>
#include <stdbool.h>
#include <stdio.h>
int main() {
int n = 29;
int cnt = 0;
// If number is less than/equal
// to 1 and number is even accept 2
// then it is not prime
if (n <= 1 || ((n > 2) && (n%2 == 0)))
printf("%d is NOT prime", n);
else {
if(n==2){
printf("%d is prime", n);
return 0;
}else{
// Check how many numbers divide n in
// range 2 to sqrt(n)
for (int i = 3; i * i <= n; i+=2) {
if (n % i == 0)
cnt++;
}
// if cnt is greater than 0 then n is
// not prime
if (cnt > 0)
printf("%d is NOT prime", n);
// else n is prime
else
printf("%d is prime", n);
}
}
return 0;
}
Output
29 is prime