CODEWITHSUNDEEP
javaloopsnumbersprimes
Dhruva
Dhruva

Reputation: 21

Prime Numbers In a given range in Java

I am trying to take the input from the user for a range and then finding all the prime numbers in that range. I am using the logic that any nuber that is greater than 2 and less than itself, which is not divisible by any number in this range (2 to less than itself) is a prime number.

public class Main
{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter the range : ");
        int n1=sc.nextInt();
        int n2=sc.nextInt();
        int fl=0;
        for(int i=n1;i<n2;i++)
        {
            for(int j=2;j<i;j++)
            { 
                if(i % j == 0)
                {
                    fl=1;
                }
            }
            if(fl == 0)
                System.out.println(i);
        }
        
    }
}

This is the output I am getting: 1 2 3

When the range is 1 to 20 (No more than 3). Please help me out.

Upvotes: 0

Views: 6631

Answers (3)

mikyll98
mikyll98

Reputation: 2283

As pointed out by Turamarth, you need to reset the flag at each iteration. I also suggest you to loop until i <= n2, otherwise you would miss the last number, and add an input check.

public static void main(String args[])
{
    int n1;
    int n2;
    boolean flag;
    
    Scanner sc = new Scanner(System.in);
    System.out.print("Enter the range: ");
    n1 = sc.nextInt();
    n2 = sc.nextInt();
    sc.close();
    
    if (n1 < 2)
        n1 = 2;
    if (n2 < 2)
    {
        System.out.println("No prime numbers in range.");
        return;
    }
    
    System.out.println("Result: ");
    for(int i = n1; i <= n2; i++)
    {
        flag = false;
        for(int j = 2; j < i; j++)
        { 
            if(i % j == 0)
            {
                flag = true;
            }
        }
        if(!flag)
            System.out.println(i);
    }
}

Example with [1, 19]:

Enter the range: 1 19
Result: 
2
3
5
7
11
13
17
19

Optimizations

Consider that:

  • you could break out from the second loop when you find a divisor (i % j == 0), by putting a break; statement inside the if condition;
  • after 2 all prime numbers are odd, therefore when i >= 3 you can increase it by 2: i += 2;
  • as suggested by user85421, instead of testing up to j < i, you can just test up to j * j <= i (equivalent to j <= sqrt(i)). In fact, there's no need to check divisors greater than the square root of i, as any larger divisor would correspond to a smaller one already tested. For example, if the number to check is 91, we just need to test up until sqrt(91) = 9.539 ~= 9, so 2, 3, 5 and 7. We don't check 13 and the others, because their corresponding divisors (in this case, for 13 is 7, because 7 * 13 = 91) would have been already checked.

Upvotes: 1

Guest123
Guest123

Reputation: 1

public static void main(String[] args){
    Scanner sc=new Scanner(System.in);
    System.out.println("enter n1:");
    int n1=sc.nextInt();
    System.out.println("enter n2:");
    int n2=sc.nextInt();
    int i;
    int j;
    int flag=0;
    for(i=n1;i<=n2;i++)
    {
    for(j=2;j<i ;j++)
    {
        if(i % j==0)
        {
            flag++;
            break;
        }
    }
    if(flag==0)
        {
         System.out.println("i= "+i);
        }
    flag=0;

    }
}

Upvotes: -1

Turamarth
Turamarth

Reputation: 2282

You have to reset your flag fl back to 0 at the start of every iteration of the outer loop.
At the moment once you encounter a non-prime (i.e. 4) it will set the value to 1 and then keep it for the remaining numbers.

Upvotes: 0

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