Program to check if given number N is Prime or not

Given a number N, the task is to check if the number is a prime number or not.

Check if given number N is Prime or not

Examples:

Input: N = 11
Output: true
Explanation: The number is not divisible by any number, other than 1 and 11 itself.

Input: N = 35
Output: false
Explanation: Apart from 1 and 35, this number is also divisible by 5 and 7.

Naive Approach (using recursion): To check the number is prime or not using recursion follow the below idea:

Recursion can also be used to check if a number between 2 to n – 1 divides n. If we find any number that divides, we return false.

Below is the implementation for the below idea:

C++

// C++ program to check whether a number
// is prime or not using recursion
#include <iostream>
using namespace std;

// function check whether a number
// is prime or not
bool isPrime(int n)
{
    static int i = 2;

    // corner cases
    if (n == 0 || n == 1) {
        return false;
    }

    // Checking Prime
    if (n == i)
        return true;

    // base cases
    if (n % i == 0) {
        return false;
    }
    i++;
    return isPrime(n);
}

// Driver Code
int main()
{
    isPrime(35) ? cout << " true\n" : cout << " false\n";
    return 0;
}

// This code is contributed by yashbeersingh42

Java

// Java program to check whether a number
// is prime or not using recursion
import java.io.*;

class GFG {

    static int i = 2;

    // Function check whether a number
    // is prime or not
    public static boolean isPrime(int n)
    {

        // Corner cases
        if (n == 0 || n == 1) {
            return false;
        }

        // Checking Prime
        if (n == i)
            return true;

        // Base cases
        if (n % i == 0) {
            return false;
        }
        i++;
        return isPrime(n);
    }

    // Driver Code
    public static void main(String[] args)
    {
        if (isPrime(35)) {
            System.out.println("true");
        }
        else {
            System.out.println("false");
        }
    }
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to check whether a number
# is prime or not using recursion

# Function check whether a number
# is prime or not


def isPrime(n, i):

    # Corner cases
    if (n == 0 or n == 1):
        return False

    # Checking Prime
    if (n == i):
        return True

    # Base cases
    if (n % i == 0):
        return False

    i += 1

    return isPrime(n, i)


# Driver Code
if (isPrime(35, 2)):
    print("true")
else:
    print("false")

# This code is contributed by bunnyram19

C#

// C# program to check whether a number
// is prime or not using recursion
using System;
class GFG {

    static int i = 2;

    // function check whether a number
    // is prime or not
    static bool isPrime(int n)
    {

        // corner cases
        if (n == 0 || n == 1) {
            return false;
        }

        // Checking Prime
        if (n == i)
            return true;

        // base cases
        if (n % i == 0) {
            return false;
        }
        i++;
        return isPrime(n);
    }

    static void Main()
    {
        if (isPrime(35)) {
            Console.WriteLine("true");
        }
        else {
            Console.WriteLine("false");
        }
    }
}

// This code is contributed by divyesh072019

Javascript

<script>
      // JavaScript program to check whether a number
      // is prime or not using recursion

      // function check whether a number
      // is prime or not
      function isPrime(n) {
        var i = 1;

        // corner cases
        if (n == 0 || n == 1) {
          return false;
        }

        // Checking Prime
        if (n == i) return true;

        // base cases
        if (n % i == 0) {
          return false;
        }
        i++;
        return isPrime(n);
      }

      // Driver Code

      isPrime(35) ? document.write(" true\n") : document.write(" false\n");

      // This code is contributed by rdtank.
    </script>

Output

 false

Time Complexity: O(N)
Auxiliary Space: O(N)

Efficient Approach: An efficient solution is :

To iterate through all numbers from 2 to sqrt(n) and for every number check if it divides n. If we find any number that divides, we return false.

Below is the implementation:

C++14

// A school method based C++ program to
// check if a number is prime
#include <bits/stdc++.h>
using namespace std;

// function check whether a number
// is prime or not
bool isPrime(int n)
{
    // Corner case
    if (n <= 1)
        return false;

    // Check from 2 to square root of n
    for (int i = 2; i <= sqrt(n); i++)
        if (n % i == 0)
            return false;

    return true;
}

// Driver Code
int main()
{
    isPrime(11) ? cout << " true\n" : cout << " false\n";
    return 0;
}

Java

// A school method based Java program to
// check if a number is prime
import java.lang.*;
import java.util.*;

class GFG {

    // Check for number prime or not
    static boolean isPrime(int n)
    {

        // Check if number is less than
        // equal to 1
        if (n <= 1)
            return false;

        // Check if number is 2
        else if (n == 2)
            return true;

        // Check if n is a multiple of 2
        else if (n % 2 == 0)
            return false;

        // If not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (n % i == 0)
                return false;
        }
        return true;
    }

    // Driver code
    public static void main(String[] args)
    {
        if (isPrime(19))
            System.out.println("true");

        else
            System.out.println("false");
    }
}

// This code is contributed by Ronak Bhensdadia

Python3

# A school method based Python3 program
# to check if a number is prime

# function check whether a number
# is prime or not

# import sqrt from math module
from math import sqrt


def isPrime(n):

    # Corner case
    if (n <= 1):
        return False

    # Check from 2 to sqrt(n)
    for i in range(2, int(sqrt(n))+1):
        if (n % i == 0):
            return False

    return True


# Driver Code
if isPrime(11):
    print("true")
else:
    print("false")

# This code is contributed by Sachin Bisht

C#

// A school method based C# program to
// check if a number is prime
using System;

class GFG {
    // function check whether a
    // number is prime or not
    static bool isPrime(int n)
    {
        // Corner case
        if (n <= 1)
            return false;

        // Check from 2 to sqrt(n)
        for (int i = 2; i < Math.Sqrt(n); i++)
            if (n % i == 0)
                return false;

        return true;
    }

    // Driver Code
    static void Main()
    {
        if (isPrime(11))
            Console.Write(" true");

        else
            Console.Write(" false");
    }
}

// This code is contributed by Sam007

PHP

<?php
// A school method based PHP program to
// check if a number is prime

// function check whether a number
// is prime or not
function isPrime($n)
{
    // Corner case
    if ($n <= 1)
        return false;

    // Check from 2 to n-1
    for ($i = 2; $i < $n; $i++)
        if ($n % $i == 0)
            return false;

    return true;
}

// Driver Code
if(isPrime(11))
    echo("true");
else
    echo("false");

// This code is contributed by Ajit.
?>

Javascript

// A school method based Javascript program to
// check if a number is prime


// function check whether a number
// is prime or not
function isPrime(n)
{
    // Corner case
    if (n <= 1)
        return false;

    // Check from 2 to n-1
    for (let i = 2; i < n; i++)
        if (n % i == 0)
            return false;

    return true;
}

// Driver Code

    isPrime(11) ? console.log(" true" + "<br>") : console.log(" false" + "<br>");

// This code is contributed by Mayank Tyagi

Output

 true

Time Complexity: O(sqrt(n))
Auxiliary space: O(1)

Another Efficient Approach: To check whether  the number is prime or not follow the below idea:

In the previous approach given if the size of the given number is too large then its square root will be also very large, so to deal with large size input we will deal with a few numbers such as 1, 2, 3, and the numbers which are divisible by 2 and 3 in separate cases and for remaining numbers, we will iterate our loop from 5 to sqrt(n) and check for each iteration whether that  (iteration) or (that iteration + 2) divides n or not. If we find any number that divides, we return false.

Below is the implementation for the above idea:

C++

// A school method based C++ program to
// check if a number is prime
#include <bits/stdc++.h>
using namespace std;

// function check whether a number
// is prime or not
bool isPrime(int n)
{
    // Check if n=1 or n=0
    if (n <= 1)
        return false;
    // Check if n=2 or n=3
    if (n == 2 || n == 3)
        return true;
    // Check whether n is divisible by 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return false;
    // Check from 5 to square root of n
    // Iterate i by (i+6)
    for (int i = 5; i <= sqrt(n); i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

// Driver Code
int main()
{
    isPrime(11) ? cout << "true\n" : cout << "false\n";
    return 0;
}
// This code is contributed by Suruchi kumari

C

// A school method based C program to
// check if a number is prime
#include <math.h>
#include <stdio.h>
// function check whether a number
// is prime or not
int isPrime(int n)
{
    // Check if n=1 or n=0
    if (n <= 1)
        return 0;
    // Check if n=2 or n=3
    if (n == 2 || n == 3)
        return 1;
    // Check whether n is divisible by 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return 0;
    // Check from 5 to square root of n
    // Iterate i by (i+6)
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return 0;

    return 1;
}

// Driver Code
int main()
{
    if (isPrime(11) == 1)
        printf("true\n");
    else
        printf("false\n");
    return 0;
}
// This code is contributed by Suruchi Kumari

Java

// Java program to check whether a number
import java.lang.*;
import java.util.*;

class GFG {

    // Function check whether a number
    // is prime or not
    public static boolean isPrime(int n)
    {
        if (n <= 1)
            return false;

        // Check if n=2 or n=3
        if (n == 2 || n == 3)
            return true;

        // Check whether n is divisible by 2 or 3
        if (n % 2 == 0 || n % 3 == 0)
            return false;

        // Check from 5 to square root of n
        // Iterate i by (i+6)
        for (int i = 5; i <= Math.sqrt(n); i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;

        return true;
    }

    // Driver Code
    public static void main(String[] args)
    {
        if (isPrime(11)) {
            System.out.println("true");
        }
        else {
            System.out.println("false");
        }
    }
}

// This code is contributed by Sayan Chatterjee

Python3

# function check whether a number
# is prime or not
from math import * def isPrime(n):

    # Check if n = 1 or n = 0
    if (n <= 1):
        return "false"

    # Check if n = 2 or n = 3
    if (n == 2 or n == 3):
        return "true"

    # Check whether n is divisible by 2 or 3
    if (n % 2 == 0 or n % 3 == 0):
        return "false"

    # Check from 5 to square root of n
       # Iterate i by (i + 6)
    for i in range(5, floor(sqrt(n)), 6):
        if (n % i == 0 or n % (i + 2) == 0):
            return "false"

    return "true"


if isPrime(11):
    print("true\n")
else:
    print("false\n")

# This code is contributed by uomkar369

C#

// C# program to check whether a number
using System;
class GFG {

    // Function check whether a number
    // is prime or not
    public static bool isPrime(int n)
    {
        if (n <= 1)
            return false;

        // Check if n=2 or n=3
        if (n == 2 || n == 3)
            return true;

        // Check whether n is divisible by 2 or 3
        if (n % 2 == 0 || n % 3 == 0)
            return false;

        // Check from 5 to square root of n
        // Iterate i by (i+6)
        for (int i = 5; i <= Math.Sqrt(n); i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;

        return true;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        if (isPrime(11)) {
            Console.WriteLine("true");
        }
        else {
            Console.WriteLine("false");
        }
    }
}

// This code is contributed by Abhijeet
// Kumar(abhijeet_19403)

Javascript

// A school method based JS program to
// check if a number is prime


// function check whether a number
// is prime or not
function isPrime(n)
{
    // Check if n=1 or n=0
    if (n <= 1)
        return false;
    // Check if n=2 or n=3
    if (n == 2 || n == 3)
        return true;
    // Check whether n is divisible by 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return false;
    // Check from 5 to square root of n
    // Iterate i by (i+6)
    for (var i = 5; i <= Math.sqrt(n); i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

// Driver Code
isPrime(11) ? console.log("true") : console.log("false");


//  This code is contributed by phasing17

Output

true

Time complexity: O(sqrt(n))
Auxiliary space: O(1)