Sum of Prime Numbers in Java

In this section, we will create Java programs to find the sum of all the prime numbers in a given range. Before moving ahead in this section, let's see the important facts about prime numbers.

A prime number is a number that is greater than 1 and can be divided by 1 and itself without leaving a remainder.
The numbers 0 and 1 are not prime numbers.
The only even prime number is 2. All other even numbers are divisible by 2.

Steps to Find the Sum of Prime Numbers

Read or initialize the lower and upper limit.
Iterate a loop (for or while) to find the prime numbers between the given range.
If the number is prime, add that number to the variable sum and print the result.

Let's implement the above steps in a Java program.

Java Program to Find the Sum of Prime Numbers

There are the following ways to find the sum of prime numbers:

Using Java for Loop
Using Java while Loop
Using Function
Using Dynamic Programming

Let's create the Java program for each.

Using for Loop

SumOfPrimeNumbersExample1.java

public class SumOfPrimeNumbers1 
{
public static void main(String args[]) 
{
int count, sum = 0;
//the loop executes 100 time and increments the variable number by 1 after each iteration
for(int number = 1; number <= 200; number++)
{
count = 0;
for (int i = 2; i <= number/2; i++)
{
//find the remainder and check if it is equal to 0 or not    
if(number % i == 0)
{
//increments the count variable by 1 if the above condition returns true
count++;
break;
} //end of if statement 
} //end of for loop
//returns true if both conditions are true
if(count == 0 && number != 1 )
{
//calculates the sum of prime numbers    
sum = sum + number;
}  //end of if statement 
}  //end of for loop
//prints the sum
System.out.println("The Sum of Prime Numbers from 1 to 200 is: " + sum);
} //end of main
} //end of class

Output:

The Sum of Prime Numbers from 1 to 200 is: 4227

Using while Loop

In the following program, we have used a while loop instead of for loop. The logic is the same as above.

SumOfPrimeNumbersExample2.java

public class SumOfPrimeNumbersExample2
{
public static void main(String args[]) 
{
int number = 1, count, sum = 0;
//executes until the condition becomes false
while(number <= 100)
{
count = 0;
int i = 2;
while(i <= number/2 )
{
//find the remainder and check if it is equal to 0 or not        
if(number % i == 0)
{
//increments the count variable by 1 if the above condition returns true
count++;
break;
}
//increments the variable i by 1
i++;
} //end of while
if(count == 0 && number != 1 )		    
{
//calculates the sum of prime numbers    
sum = sum + number;
} //end of if statement
//increments the variable number by 1 after completing each iteration
number++;
}  //end of while
//prints the sum
System.out.println("The Sum of Prime Numbers from 1 to 100 is: " + sum);
}  //end of main
} //end of class

Output:

The Sum of Prime Numbers from 1 to 100 is: 1060

Using Function

SumOfPrimeNumbersExample3.java

public class SumOfPrimeNumbersExample3
{
//function that checks if the given number is prime or not
static boolean isPrime(int num)
{
if(num == 1) 
{
return false;
} //end of if
for (int i = 2; i*i <= num; i++) 
{
if (num % i == 0) 
{
return false;
} //end of if statement
} //end of for
return true;
}
//method iterate over min to max
//function that calculates the sum, if the number is detected as prime
static int addPrime(int min, int max)
{
int sum = 0;
//loop starts from 50 and decrement the variable i by 1 after each iteration
for (int i = max; i >= min; i--) 
{
//calling the user-defined method to check the number is prime or not
boolean prime = isPrime(i);
if (prime) 
{
//calculates the sum 
sum = sum + i;
} //end of if statement
} //end of for loop
//returns sum
return sum;
}
//driver code
public static void main(String args[])
{
int min = 0, max = 50;
//call the user-defined method and print the sum
System.out.println("The sum of all the prime number between the given range is: " +addPrime(min, max));
} //end of main
} //end of class

Output:

The sum of all the prime number between the given range is: 328

Using Dynamic Programming

The logic for the sum of prime numbers using dynamic programming is a bit tricky and difficult to implement. In this approach, we will use arrays. Let's see the steps.

Declare two arrays number[] and array[].
Initialize array[] to 0.
Iterate over the for loop till sqrt(uprlimit). If the array element is 0, consider it as prime and its multiple as non-prime. While marking the non-prime numbers, set the respective location to 1.
Update the number[] array that holds the sum of prime numbers. Each position of number[i] represents the sum of all prime numbers within a specified range i.e. [1, i].

Let's implement the above steps in a Java program.

SumOfPrimeNumbersExample4.java

public class SumOfPrimeNumbersExample4
{
//suppose the constraint is N<=1000
static int uprlimit = 1000;
//declaring an array
static long numbers[] = new long[uprlimit + 1];
//Method to compute the array
static void findSum()
{
//declaring another array
int array[] = new int[uprlimit + 1];
//initialize 0th and 1st element of the array to 1
array[0] = 1; 
array[1] = 1;
//2 is an even prime number so we will iterate the loop from 2 to sqrt(uprlimit)
for (int i = 2; i <= Math.sqrt(uprlimit); i++)
//if ith element of the array is 0, mark it as prime
if (array[i] == 0)
//mark all of it's multiples till uprlimit as non-prime 
for (int j = i * i; j <= uprlimit; j = j + i)
array[j] = 1;
long primeSum = 0;
//update the array number[] with the running sum of prime numbers within the range [1, uprlimt]
for (int i = 1; i <= uprlimit; i++)
{
if (array[i] == 0)
primeSum = primeSum + i;
//here, number[i] represents the sum of all the prime numbers within the range [1, i]
numbers[i] = primeSum;
} //end of for
} //end of method
//driver code
public static void main(String args[])
{
int lr = 100, ur = 1000;
//calling user-defined function
findSum();
System.out.println("The sum of all the prime number between given range is: "+(numbers[ur] - numbers[lr - 1]));
}   //end of main
}  //end of class

Output:

The sum of all the prime number between the given range is: 75067

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