**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 = 3Output:PrimeExplanation:3 is a prime numberExample 2:Input:N = 26Output:Non-PrimeExplanation: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 (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 (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 (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 (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)**

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