Program to find Prime Numbers Between given Interval

Given two numbers a and b as interval range, the task is to find the prime numbers in between this interval.

Examples:

Input : a = 1, b = 10
Output : 2, 3, 5, 7
Input : a = 10, b = 20
Output : 11, 13, 17, 19

In the below program, the range of numbers is taken as input and stored in the variables ‘a’ and ‘b’. Then using a for-loop, the numbers between the interval of a and b are traversed. For each number in the for loop, it is checked if this number is prime or not. If found prime, print the number. Then the next number in the loop is checked, till all numbers are checked.

Program:

C++

// C++ program to find the prime numbers
// between a given interval
 
#include <bits/stdc++.h>
using namespace std;
 
int main()
{
    // Declare the variables
    int a, b, i, j, flag;
 
    // Ask user to enter lower value of interval
    cout << "Enter lower bound of the interval: ";
    cin >> a; // Take input
 
    // Ask user to enter upper value of interval
    cout << "\nEnter upper bound of the interval: ";
    cin >> b; // Take input
 
    // Print display message
    cout << "\nPrime numbers between "
         << a << " and " << b << " are: ";
 
    // Traverse each number in the interval
    // with the help of for loop
    for (i = a; i <= b; i++) {
        // Skip 0 and 1 as they are
        // neither prime nor composite
        if (i == 1 || i == 0)
            continue;
 
        // flag variable to tell
        // if i is prime or not
        flag = 1;
 
        for (j = 2; j <= i / 2; ++j) {
            if (i % j == 0) {
                flag = 0;
                break;
            }
        }
 
        // flag = 1 means i is prime
        // and flag = 0 means i is not prime
        if (flag == 1)
            cout << i << " ";
    }
 
    return 0;
}
 
// This code is contributed by Akanksha Rai

C

// C program to find the prime numbers
// between a given interval
 
#include <stdio.h>
 
int main()
{
    // Declare the variables
    int a, b, i, j, flag;
 
    // Ask user to enter lower value of interval
    printf("Enter lower bound of the interval: ");
    scanf("%d", &a); // Take input
 
    // Ask user to enter upper value of interval
    printf("\nEnter upper bound of the interval: ");
    scanf("%d", &b); // Take input
 
    // Print display message
    printf("\nPrime numbers between %d and %d are: ", a, b);
 
    // Traverse each number in the interval
    // with the help of for loop
    for (i = a; i <= b; i++) {
        // Skip 0 and 1 as they are
        // neither prime nor composite
        if (i == 1 || i == 0)
            continue;
 
        // flag variable to tell
        // if i is prime or not
        flag = 1;
 
        for (j = 2; j <= i / 2; ++j) {
            if (i % j == 0) {
                flag = 0;
                break;
            }
        }
 
        // flag = 1 means i is prime
        // and flag = 0 means i is not prime
        if (flag == 1)
            printf("%d ", i);
    }
 
    return 0;
}

Java

import java.util.Scanner;
 
// Java program to find the prime numbers
// between a given interval
public class GFG {
 
    // driver code
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        // Declare the variables
        int a, b, i, j, flag;
 
        // Ask user to enter lower value of interval
        System.out.printf("Enter lower bound of the interval: ");
        a = sc.nextInt(); // Take input
 
        // Ask user to enter upper value of interval
        System.out.printf("\nEnter upper bound of the interval: ");
        b = sc.nextInt(); // Take input
 
        // Print display message
        System.out.printf("\nPrime numbers between %d and %d are: ", a, b);
 
        // Traverse each number in the interval
        // with the help of for loop
        for (i = a; i <= b; i++) {
 
            // Skip 0 and 1 as they are
            // neither prime nor composite
            if (i == 1 || i == 0)
                continue;
 
            // flag variable to tell
            // if i is prime or not
            flag = 1;
 
            for (j = 2; j <= i / 2; ++j) {
                if (i % j == 0) {
                    flag = 0;
                    break;
                }
            }
 
            // flag = 1 means i is prime
            // and flag = 0 means i is not prime
            if (flag == 1)
                System.out.println(i);
        }
    }
}

Python3

# Python3 program to find the prime
# numbers between a given interval
 
if __name__ == '__main__':
     
    # Declare the variables
    a, b, i, j, flag = 0, 0, 0, 0, 0
 
    # Ask user to enter lower value of interval
    print("Enter lower bound of the interval:",
                                      end = "")
    a = int(input()) # Take input
    print(a)
     
    # Ask user to enter upper value of interval
    print("Enter upper bound of the interval:",
                                      end = "")
    b = int(input()) # Take input
    print(b)
     
    # Print display message
    print("Prime numbers between", a, "and",
                        b, "are:", end = "")
 
    # Traverse each number in the interval
    # with the help of for loop
    for i in range(a, b + 1):
 
        # Skip 1 as1 is neither
        # prime nor composite
        if (i == 1):
            continue
 
        # flag variable to tell
        # if i is prime or not
        flag = 1
         
        for j in range(2, i // 2 + 1):
            if (i % j == 0):
                flag = 0
                break
             
        # flag = 1 means i is prime
        # and flag = 0 means i is not prime
        if (flag == 1):
            print(i, end = " ")
             
# This code is contributed
# by Mohit kumar 29

C#

// C# program to find the prime numbers
// between a given interval
using System;
 
class GFG{
 
// Driver code   
public static void Main(string[] args)
{
     
    // Declare the variables
    int a, b, i, j, flag;
 
    // Ask user to enter lower value of interval
    Console.WriteLine("Enter lower bound of " +
                      "the interval: ");
                       
    // Take input
    a = int.Parse(Console.ReadLine());
 
    // Ask user to enter upper value of interval
    Console.WriteLine("\nEnter upper bound " +
                      "of the interval: ");
                       
    // Take input
    b = int.Parse(Console.ReadLine());
 
    // Print display message
    Console.WriteLine("\nPrime numbers between " + 
                      "{0} and {1} are: ", a, b);
 
    // Traverse each number in the interval
    // with the help of for loop
    for(i = a; i <= b; i++)
    {
         
        // Skip 0 and 1 as they are
        // neither prime nor composite
        if (i == 1 || i == 0)
            continue;
 
        // flag variable to tell
        // if i is prime or not
        flag = 1;
 
        for(j = 2; j <= i / 2; ++j)
        {
            if (i % j == 0)
            {
                flag = 0;
                break;
            }
        }
 
        // flag = 1 means i is prime
        // and flag = 0 means i is not prime
        if (flag == 1)
            Console.WriteLine(i);
    }
}
}
 
// This code is contributed by jana_sayantan

Javascript

<script>
 
// JavaScript program to find the prime numbers
// between a given interval
 
// driver code
    // Declare the variables
    let a, b, i, j, flag;
 
    // Ask user to enter lower value of interval and take input
    a = window.prompt("Enter lower bound of the interval: ");
 
    // Ask user to enter upper value of interval and take input
    b = window.prompt("Enter upper bound of the interval: ");
 
    // Print display message
    console.log("Prime numbers between " + a + " and " + b << " are: ");
 
    // Traverse each number in the interval
    // with the help of for loop
    for (i = a; i <= b; i++) {
        // Skip 0 and 1 as they are
        // neither prime nor composite
        if (i == 1 || i == 0)
            continue;
 
        // flag variable to tell
        // if i is prime or not
        flag = 1;
 
        for (j = 2; j <= i / 2; ++j) {
            if (i % j == 0) {
                flag = 0;
                break;
            }
        }
 
        // flag = 1 means i is prime
        // and flag = 0 means i is not prime
        if (flag == 1)
            document.write(i," ");
    }
 
// This code is contributed by shinjanpatra
 
</script>

Output:

Enter lower bound of the interval: 1
Enter upper bound of the interval: 10
Prime numbers between 1 and 10 are: 2 3 5 7

Time Complexity: O(N²), Where N is the difference between the range
Auxiliary Space: O(1)

Optimized Solution :

The idea is to use the fact that even numbers (except 2) are not primes.

C++

// C++ program to find the prime numbers
// between a given interval
 
#include <bits/stdc++.h>
using namespace std;
 
int main()
{
    // Declare the variables
    int a, b, i, j;
 
    // Ask user to enter lower value of interval
    // interval < 0 cannot be prime numbers
    cout << "Enter lower bound of the interval: ";
    cin >> a; // Take input
 
    // Ask user to enter upper value of interval
    cout << "\nEnter upper bound of the interval: ";
    cin >> b; // Take input
 
    // Print display message
    cout << "\nPrime numbers between " << a << " and " << b
         << " are: ";
 
    // Explicitly handling the cases when a is less than 2
    // since 0 and 1 are not prime numbers
    if (a <= 2) {
        a = 2;
        if (b >= 2) {
            cout << a << " ";
            a++;
        }
    }
 
    // MAKING SURE THAT a IS ODD BEFORE WE BEGIN
    // THE LOOP
    if (a % 2 == 0)
        a++;
 
    // NOTE : WE TRAVERSE THROUGH ODD NUMBERS ONLY
    for (i = a; i <= b; i = i + 2) {
 
        // flag variable to tell
        // if i is prime or not
        bool flag = 1;
 
        // WE TRAVERSE TILL SQUARE ROOT OF j only.
        // (LARGEST POSSIBLE VALUE OF A PRIME FACTOR)
        for (j = 2; j * j <= i; ++j) {
            if (i % j == 0) {
                flag = 0;
                break;
            }
        }
 
        // flag = 1 means i is prime
        // and flag = 0 means i is not prime
       
        if (flag == 1){
          if(i==1)
            continue;
          else
            cout << i << " ";
        }
    }
 
    return 0;
}

Java

import java.util.*;
 
public class PrimeNumbersInRange {
 
    // function which checks whether a number is Prime or Not
    // If the number is prime, it returns true. Else, it returns false.
    public static boolean isPrime(int n) {
        // 0 and 1 are neither prime nor composite numbers
        if (n == 0 || n == 1) {
            return false;
        }
        // 2 is a prime number
        if (n == 2) {
            return true;
        }
        // every composite number has a prime factor
        // less than or equal to its square root.
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
 
    }
 
    // Driver code
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
 
        // Ask user to enter lower value of interval
        System.out.print("Enter lower bound of the interval: ");
        int lower = sc.nextInt(); // Take input
 
        // Ask user to enter upper value of interval
        System.out.print("\nEnter upper bound of the interval: ");
        int upper = sc.nextInt(); // Take input
 
        // Print display message
        System.out.printf("\nPrime numbers between %d and %d are: ", lower, upper);
 
        for (int i = lower; i <= upper; i++) {
            if (isPrime(i)) {
                System.out.print(i + " ");
            }
        }
        sc.close();
    }
}

Python3

# Python3 program to find the prime
# numbers between a given interval
  
if __name__ == '__main__':
     
    # Declare the variables
    a, b, i, j = 0, 0, 0, 0
     
    # Ask user to enter lower value of interval
    print("Enter lower bound of the interval:",end = "")
     
    a = int(input()) # Take input
    print(a)
     
    # Ask user to enter upper value of interval
    print("Enter upper bound of the interval:",end = "")
     
    b = int(input()) # Take input
    print(b)
     
    # Print display message
    print("Prime numbers between", a, "and",b, "are:", end = "")
     
     
    # Explicitly handling the cases when a is less than 2
    if (a == 1):
        print(a,end=" ")
        a+=1
        if (b >= 2):
            print(a,end=" ")
            a+=1
    if (a == 2):
        print(a,end=" ")
     
    # MAKING SURE THAT a IS ODD BEFORE WE BEGIN
    # THE LOOP
    if (a % 2 == 0):
        a+=1
    # NOTE : WE TRAVERSE THROUGH ODD NUMBERS ONLY
    for i in range(a,b+1,2):
         
        # flag variable to tell
        # if i is prime or not
        flag = 1
        # WE TRAVERSE TILL SQUARE ROOT OF j only.
        # (LARGEST POSSIBLE VALUE OF A PRIME FACTOR)
        j = 2
        while(j * j <= i):
            if (i % j == 0):
                flag = 0
                break
            j+=1
         
        # flag = 1 means i is prime
        # and flag = 0 means i is not prime
        if (flag == 1):
            print(i,end=" ")
 
# This code is contributed by shubhamsingh10

C#

// C# program to find the prime numbers
// between a given interval
using System;
class GFG
{
  // Driver code
  static public void Main()
  {
 
    // Declare the variables
    int a, b, i, j, flag;
 
    // Ask user to enter lower value of interval
    Console.Write(
      "Enter lower bound of the interval: ");
    a = Convert.ToInt32(
      Console.ReadLine()); // Take input
 
    // Ask user to enter upper value of interval
    Console.Write(
      "\nEnter upper bound of the interval: ");
    b = Convert.ToInt32(
      Console.ReadLine()); // Take input
 
    // Print display message
    Console.Write("\nPrime numbers between " + a
                  + " and " + b + " are: ");
 
    // Explicitly handling the cases when a is less than
    // 2
    if (a == 1)
    {
      Console.Write(a + " ");
      a++;
      if (b >= 2)
      {
        Console.Write(a + " ");
        a++;
      }
    }
    if (a == 2)
    {
      Console.Write(a + " ");
    }
 
    // MAKING SURE THAT a IS ODD BEFORE WE BEGIN
    // THE LOOP
    if (a % 2 == 0)
    {
      a++;
    }
 
    // NOTE : WE TRAVERSE THROUGH ODD NUMBERS ONLY
    for (i = a; i <= b; i = i + 2)
    {
 
      // flag variable to tell
      // if i is prime or not
      flag = 1;
 
      // WE TRAVERSE TILL SQUARE ROOT OF j only.
      // (LARGEST POSSIBLE VALUE OF A PRIME FACTOR)
      for (j = 2; j * j <= i; ++j)
      {
        if (i % j == 0)
        {
          flag = 0;
          break;
        }
      }
 
      // flag = 1 means i is prime
      // and flag = 0 means i is not prime
      if (flag == 1)
      {
        Console.Write(i + " ");
      }
    }
  }
}
 
// This code is contributed by rag2127

Javascript

<script>
 
// JavaScript program to find the prime numbers
// between a given interval
 
// driver code
    // Declare the variables
    let a, b, i, j;
 
    // Ask user to enter lower value of interval and take input
    a = window.prompt("Enter lower bound of the interval: ");
 
    // Ask user to enter upper value of interval and take input
    b = window.prompt("Enter upper bound of the interval: ");
 
    // Print display message
    console.log("Prime numbers between " + a + " and " + b << " are: ");
 
    // Explicitly handling the cases when a is less than 2
    // since 0 and 1 are not prime numbers
    if (a <= 2)
    {
        a = 2;
        if (b >= 2)
        {
            document.write(a," ");
            a++;
        }
    }
 
    // MAKING SURE THAT a IS ODD BEFORE WE BEGIN
    // THE LOOP
    if (a % 2 == 0)
        a++;
 
    // NOTE : WE TRAVERSE THROUGH ODD NUMBERS ONLY
    for (i = a; i <= b; i = i + 2) {
 
        // flag variable to tell
        // if i is prime or not
        let flag = 1;
 
        // WE TRAVERSE TILL SQUARE ROOT OF j only.
        // (LARGEST POSSIBLE VALUE OF A PRIME FACTOR)
        for (j = 2; j * j <= i; ++j) {
            if (i % j == 0) {
                flag = 0;
                break;
            }
        }
 
        // flag = 1 means i is prime
        // and flag = 0 means i is not prime
       
        if (flag == 1){
          if(i == 1)  continue;
          else
            document.write(i, " ");
        }
    }
 
// This code is contributed by shinjanpatra
</script>

Output:

Enter lower bound of the interval: 1
Enter upper bound of the interval: 10
Prime numbers between 1 and 10 are: 2 3 5 7

Time Complexity: O(N*sqrt(N)), Where N is the difference between the range
Auxiliary Space: O(1)

Sieve of Eratosthenes Method:

The sieve of Eratosthenes is one of the most efficient ways to find all primes smaller than N when N is smaller than 10 million or so.

Steps:

Create a boolean array prime[srt to n] and initialize all its entries as true.
Mark prime[0] and prime[1] as false because they are not prime numbers.
Starting from p = srt, for each prime number p less than or equal to the square root of n, mark all multiples of p greater than or equal to p*p as composite by setting prime[i] to false.
Finally, print all prime numbers between srt and n.

Below is the implementation:

C++

// C++ program to find the prime numbers
// between a given interval using Sieve of Eratosthenes
#include <bits/stdc++.h>
using namespace std;
 
void SieveOfEratosthenes(int srt, int n)
{
    // Create a boolean array "prime[srt to n]" and
    // initialize all entries it as true. A value in
    // prime[i] will finally be false if i is Not a prime,
    // else true.
    bool prime[n + 2];
    memset(prime, true, sizeof(prime));
    prime[0] = false;
    prime[1] = false;
 
    for (int p = 2; p * p <= n; p++) {
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true) {
            // Update all multiples of p greater than or
            // equal to the square of it numbers which are
            // multiple of p and are less than p^2 are
            // already been marked.
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
 
    // Print all prime numbers
    for (int p = srt; p <= n; p++)
        if (prime[p])
            cout << p << " ";
}
 
// Driver Code
int main()
{
    int srt = 1;
    int end = 10;
    SieveOfEratosthenes(srt, end);
    return 0;
}
 
// This code is contributed by Susobhan Akhuli

Java

// java program to find the prime numbers
// between a given interval using Sieve of Eratosthenes
import java.util.*;
 
public class GFG {
    public static void SieveOfEratosthenes(int srt, int n)
    {
        // Create a boolean array "prime[srt to n]" and
        // initialize all entries it as true. A value in
        // prime[i] will finally be false if i is Not a
        // prime, else true.
        boolean[] prime = new boolean[n + 2];
        Arrays.fill(prime, true);
        prime[0] = false;
        prime[1] = false;
 
        for (int p = 2; p * p <= n; p++) {
            // If prime[p] is not changed, then it is a
            // prime
            if (prime[p] == true) {
                // Update all multiples of p greater than or
                // equal to the square of it numbers which
                // are multiple of p and are less than p^2
                // are already been marked.
                for (int i = p * p; i <= n; i += p) {
                    prime[i] = false;
                }
            }
        }
 
        // Print all prime numbers
        for (int p = srt; p <= n; p++) {
            if (prime[p]) {
                System.out.print(p + " ");
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int srt = 1;
        int end = 10;
        SieveOfEratosthenes(srt, end);
    }
}
 
// This code is contributed by Susobhan Akhuli

Python3

# Python program to find the prime numbers
# between a given interval using Sieve of Eratosthenes
import math
 
def SieveOfEratosthenes(srt, n):
    # Create a boolean array "prime[srt to n]" and
    # initialize all entries it as true. A value in
    # prime[i] will finally be false if i is Not a prime,
    # else true.
    prime = [True for i in range(n + 2 )]
    prime[0] = False
    prime[1] = False
 
    for p in range(2, int(math.sqrt(n))+1):
        # If prime[p] is not changed, then it is a prime
        if prime[p] == True:
            # Update all multiples of p greater than or
            # equal to the square of it numbers which are
            # multiple of p and are less than p^2 are
            # already been marked.
            for i in range(p*p, n+1, p):
                prime[i] = False
 
    # Print all prime numbers
    for p in range(srt, n+1):
        if prime[p]:
            print(p, end=" ")
 
# Driver Code
if __name__ == "__main__":
    srt = 1
    end = 10
    SieveOfEratosthenes(srt, end)
 
# This code is contributed by Susobhan Akhuli

C#

// C# program to find the prime numbers
// between a given interval using Sieve of Eratosthenes
using System;
 
public class GFG
{
    public static void SieveOfEratosthenes(int srt, int n)
    {
        // Create a boolean array "prime[srt to n]" and
        // initialize all entries it as true. A value in
        // prime[i] will finally be false if i is Not a
        // prime, else true.
        bool[] prime = new bool[n + 2];
        Array.Fill(prime, true);
        prime[0] = false;
        prime[1] = false;
 
        for (int p = 2; p * p <= n; p++)
        {
            // If prime[p] is not changed, then it is a
            // prime
            if (prime[p] == true)
            {
                // Update all multiples of p greater than or
                // equal to the square of it. Numbers which
                // are multiples of p and are less than p^2
                // are already marked.
                for (int i = p * p; i <= n; i += p)
                {
                    prime[i] = false;
                }
            }
        }
 
        // Print all prime numbers
        for (int p = srt; p <= n; p++)
        {
            if (prime[p])
            {
                Console.Write(p + " ");
            }
        }
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int srt = 1;
        int end = 10;
        SieveOfEratosthenes(srt, end);
    }
}
 
// This code is contributed by Pushpesh Raj

Javascript

function SieveOfEratosthenes(srt, n) {
    // Create a boolean array "prime[srt to n]" and
    // initialize all entries it as true. A value in
    // prime[i] will finally be false if i is Not a prime,
    // else true.
    let prime = new Array(n + 2 ).fill(true);
    prime[0] = false;
    prime[1] = false;
 
    for (let p = 2; p * p <= n; p++) {
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true) {
            // Update all multiples of p greater than or
            // equal to the square of it numbers which are
            // multiple of p and are less than p^2 are
            // already been marked.
            for (let i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
 
    // Print all prime numbers
    for (let p = srt; p <= n; p++)
        if (prime[p])
            console.log(p + " ");
}
 
let srt = 1;
let end = 10;
SieveOfEratosthenes(srt, end);

Output

2 3 5 7

Time Complexity:
The time complexity of the Sieve of Eratosthenes algorithm is O(n*log(log(n))) as it iterates over all numbers from 2 to n and removes all the multiples of each prime number.

In this code, the algorithm is applied only to a range [srt, n], which reduces the time complexity to O((n-srt+1)*log(log(n))) for the range [srt, n]. The loop for marking the multiples of each prime number iterates from p*p to n, which takes O((n-srt+1)*log(log(n))) time. Therefore, the overall time complexity of this code is O((n-srt+1)*log(log(n))).

Auxiliary Space:
The space complexity of the algorithm is O(n), which is the size of the boolean array prime. However, the size of the array is reduced to n+2-srt, which is the size of the array required for the given range [srt, n]. Therefore, the auxiliary space complexity of the code is O(n-srt+1).

Last Updated : 06 Jul, 2023