# Check if a number is prime or not

Problem Statement: Given a number, check whether it is prime or not. A prime number is a natural number that is only divisible by 1 and by itself.

Examples 1 2 3 5 7 11 13 17 19 …

Examples:

```Example 1:
Input: N = 3
Output: Prime
Explanation: 3 is a prime number

Example 2:
Input: N = 26
Output: Non-Prime
Explanation: 26 is not prime```

### Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Solution 1: Using Iterative solution

Approach

A prime number is a natural number that is only divisible by 1 and by itself. Examples 1 2 3 5 7 11 13 17 19 …

Running a for loop for checking if the number is divisible by a number from 2 to a number less than a given number.

And then checking if the number is divisible by the numbers from 2 to the number less than a given number

Then, If the remainder is zero, that means it is divisible and hence not a prime number.

If the loop runs till square root and none of the numbers divided it completely. So it is the Prime number. Code:

## C++ Code

``````#include<bits/stdc++.h>

using namespace std;

bool isPrime(int N) {
for (int i = 2; i < N; i++) {
if (N % i == 0) {
return false;
}
}
return true;
}
int main() {

int n = 11;

bool ans = isPrime(n);
if (n != 1 && ans == true) {
cout << "Prime Number";
} else {
cout << "Non Prime Number";
}
return 0;
}
``````

Output: Prime Number

Time Complexity: O(n)

Space Complexity: O(1)

## Java Code

``````class Solution {

public static boolean isPrime(int N) {

for (int i = 2; i < N; i++) {
if (N % i == 0) {
return false;
}
}
return true;

}
public static void main(String args[]) {
int n = 20;
boolean ans = (isPrime(n));
if (n != 1 && ans == true) {
System.out.println("Prime Number");
} else {
System.out.println("Non-Prime Number");
}
}
}
``````

Output: Non-Prime Number

Time Complexity: O(n)

Space Complexity: O(1)

Solution 2: Optimized Approach

Approach: Running the for loop till the square root of the number

A prime number is a natural number that is only divisible by 1 and by itself. Examples 1 2 3 5 7 11 13 17 19 …

Using a for loop for checking if the number is divisible by a number from 2 to its square root.

Running the for loop from 2 to the square root of the number.

And then checking if the number is divisible by the numbers from 2 to its square root.

Then, If the remainder is zero, that means it is divisible and hence not a prime number.

If the loop runs till square root and none of the numbers divided it completely. So it is the Prime number.

Code:

## C++ Code

``````#include<bits/stdc++.h>

using namespace std;

bool isPrime(int N) {
for (int i = 2; i < sqrt(N); i++) {
if (N % i == 0) {
return false;
}
}
return true;
}
int main() {

int n = 11;

bool ans = isPrime(n);
if (n != 1 && ans == true) {
cout << "Prime Number";
} else {
cout << "Non Prime Number";
}
return 0;
}
``````

Output: Prime Number

Time Complexity: O(√n)

Space Complexity: O(1)

## Java Code

``````class Solution {

public static boolean isPrime(int N) {

for (int i = 2; i < Math.sqrt(N); i++) {
if (N % i == 0) {
return false;
}
}
return true;

}
public static void main(String args[]) {
int n = 20;
boolean ans = (isPrime(n));
if (n != 1 && ans == true) {
System.out.println("Prime Number");
} else {
System.out.println("Non Prime Number");
}
}
}
``````

Output: Non Prime Number

Time Complexity: O(√n)

Space Complexity: O(1)

Special thanks to Rushikesh Adhav for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article