C++ Program to Check Prime Number

Last Updated : 05 May, 2025

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

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.

There are many approaches for checking whether the given number is a prime number:

Table of Content

Brute Force Method – O(n) Time

In this approach, we can check whether the number is prime or not by iterating in the range from 1 to n. 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.

C++

//Driver Code Starts{ #include <bits/stdc++.h> using namespace std;  int main() { //Driver Code Ends }      int n = 29;     int cnt = 0;          // If number is less than/equal to 1,     // it is not prime     if (n <= 1)         cout << n << " is NOT prime";     else {          // Count the divisors of n         for (int i = 1; i <= n; i++) {             if (n % i == 0)                 cnt++;         }          // If n is divisible by more than 2          // numbers then it is not prime         if (cnt > 2)             cout << n << " is NOT prime";          // else it is prime         else             cout << n << " is prime";     }  //Driver Code Starts{     return 0; } //Driver Code Ends }

Output

29 is prime

The time complexity of the above program is O(n) because we iterate from 1 to n.

Optimized Method – O(√n) Time

In this approach, we don’t check all divisors of the given number, we only check divisors up to the square root of the number because, “The smallest factor of a number greater than one cannot be greater than the square root of that number.”

C++

//Driver Code Starts{ #include <bits/stdc++.h> using namespace std;  int main() { //Driver Code Ends }      int n = 29;     int cnt = 0;          // If number is less than/equal to 1,     // it is not prime     if (n <= 1)         cout << n << " is NOT prime";     else {          // Count the divisors of n         for (int i = 2; i * i <= n; i++) {             if (n % i == 0)                 cnt++;         }          // If n is divisible by more than 2          // numbers then it is not prime         if (cnt > 0)             cout << n << " is NOT prime";          // else it is prime         else             cout << n << " is prime";     }  //Driver Code Starts{     return 0; } //Driver Code Ends }

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.

C++

//Driver Code Starts{ #include <bits/stdc++.h> using namespace std;  int main() { //Driver Code Ends }      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)))         cout << n << " is NOT prime";     else {         if (n == 2) {             cout << n << " is prime";             return 0;         } else {              // Check how many numbers divide              // n in the 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)                 cout << n << " is NOT prime";              // else n is prime             else                 cout << n << " is prime";         }     }  //Driver Code Starts{     return 0; } //Driver Code Ends }

Output

29 is prime

C++ Program To Find Prime Numbers Between Given Interval

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C++ Program to Check Prime Number

Brute Force Method – O(n) Time

Optimized Method – O(√n) Time

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