Prime Number problem solve

Prime Number problem solve

 

Hello Programmers, 

Today i write a article where you can find  prime number in a array. This is a easy problem. We are know prime number start 2. So we need to first  is input number (n>1). If n is greeter then 1. Then i check is it 2. If it is 2 then we print this is prime number.   Then you check is input number divided  by 3,4,5,6........ 

if input number divided one number then we can print this is not prime number. Otherwise we print this is prime number. 

So solution is : C++ and JavaScript

// Solution 01 

// return 0 means false

// return 1 means true

 

//C++ Solution

void Prime_Num (int n){

if(n < 2){

cout<< 0 <<endl;

}else if(n == 2){

cout<< 1 <<endl;

}else {

int i;

for(i = 3; i < n; i++){

if(n % i == 0){

cout<< 0 <<endl;

break;

}

cout<< 1 <<endl;

}

}

}

//JavaScript Solution

 

function is_prime(n){

if(n < 2){

return 0;

}else if(n == 2){

return 1;

}else {

var i;

for(i = 3; i < n ; i++){

if(n % i == 0){

return 0;

break;

}

return 1;

}

}

}

We can solve this problem another way. Where we divided  input number by 2. Because if a number divided by 2 or even number then we can print this is not prime number. So we can run a loop only odd number. This is more effective solution number 1. And this is run more first. Because we don't need  check every even and odd number. 

So Solution number 2 is: C++ and JavaScriipt

 

// C++ Solution

 

void Prime_Num (int n){

if(n < 2){

cout<< 0 <<endl;

}else if(n == 2){

cout<< 1 <<endl;

}else if(n % 2 == 0){

cout<< 0 <<endl;

}else {

int i;

for(i = 3; i < n; i+2){

if(n % i == 0){

cout<< 0 <<endl;

break;

}

cout<< 1 <<endl;

}

}

}

 

 

// JavaScript Solution

function is_prime(n){

if(n < 2){

return 0;

}else if(n == 2){

return 1;

}else if(n % 2 == 0){

return 0;

}else {

var i;

for(i = 3; i < n; i+2){

if(n % i == 0){

return 0;

break;

}

return 1;

}

}

}