Find difference between count of Prime and Composite in given Range

Given two integers L and R, the task is to find the value of (number of composites – number of primes) that lie in the range [L, R].

Note: 1 is considered neither a prime nor a composite number.

Examples:

Input: L = 2, R = 10
Output: 1
Explanation: The composite numbers in the range are 4, 6, 8, 9, 10 and the prime numbers are 2, 3, 5, 7. So the output is 1.

Input: L = 4, R = 5
Output: 0

Approach: The problem can be solved by using the following idea:

Find the count of primes and the count of composites within the given range and then find the difference.

Follow the steps mentioned below to implement the idea:

• Traverse from i = L to R:
  • Traverse from j = 2 to the square root of the i.
    • If i is divisible by j, then it is not a prime number. So increment the count of composite numbers.
    • Otherwise, increment the count of prime numbers.
• Return the difference as the required answer. 

Below is the implementation of the above approach.

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the required difference
int findDiff(int L, int R)
{
    int prime = 0, comp = 0;
    bool flag;

    // Loop to iterate from L to R
    for (int i = L; i <= R; i++) {
        flag = false;

        // Loop to check if i is prime
        for (int j = 2; j <= sqrt(i); j++) {
            if (i % j == 0) {
                flag = true;
                break;
            }
        }
        if (flag)
            comp++;
        else
            prime++;
    }

    // Return the difference
    return comp - prime;
}

// Driver code
int main()
{
    int L = 2, R = 10;

    // Function call
    cout << findDiff(L, R);

    return 0;
}

Java

// Java code to implement the approach
import java.io.*;

class GFG {
    // Function to find the required difference
    public static int findDiff(int L, int R)
    {
        int prime = 0, comp = 0;
        boolean flag = true;

        // Loop to iterate from L to R
        for (int i = L; i <= R; i++) {
            flag = false;

            // Loop to check if i is prime
            for (int j = 2; j <= Math.sqrt(i); j++) {
                if (i % j == 0) {
                    flag = true;
                    break;
                }
            }
            if (flag == true)
                comp++;
            else
                prime++;
        }

        // Return the difference
        return comp - prime;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int L = 2, R = 10;

        // Function call
        System.out.print(findDiff(L, R));
    }
}

// This code is contributed by Rohit Pradhan

Python3

# Python code to implement the approach
import math

# Function to find the required difference
def findDiff(L, R):
    prime = 0
    comp = 0
    flag = True
    
    # Loop to iterate from L to R
    for i in range(L, R + 1):
        flag = False
        
        # Loop to check if i is prime
        for j in range(2,(int)(math.sqrt(i)) + 1):
            if(i%j == 0):
                flag = True
                break
        if(flag):
            comp = comp + 1
        else:
            prime = prime + 1
    
    # Return the difference
    return comp - prime
    
# Driver code
L = 2
R = 10

# Function call
print(findDiff(L, R))

# This code is contributed by Pushpesh Raj.

C#

// C# code to implement the approach
using System;

class GFG {

  // Function to find the required difference
  static int findDiff(int L, int R)
  {
    int prime = 0, comp = 0;
    bool flag = true;

    // Loop to iterate from L to R
    for (int i = L; i <= R; i++) {
      flag = false;

      // Loop to check if i is prime
      for (int j = 2; j <= Math.Sqrt(i); j++) {
        if (i % j == 0) {
          flag = true;
          break;
        }
      }
      if (flag)
        comp++;
      else
        prime++;
    }

    // Return the difference
    return comp - prime;
  }

  // Driver code
  public static void Main()
  {
    int L = 2, R = 10;

    // Function call
    Console.Write(findDiff(L, R));
  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

        // JavaScript code to implement the approach

        // Function to find the required difference
        const findDiff = (L, R) => {
            let prime = 0, comp = 0;
            let flag;

            // Loop to iterate from L to R
            for (let i = L; i <= R; i++) {
                flag = false;

                // Loop to check if i is prime
                for (let j = 2; j <= parseInt(Math.sqrt(i)); j++) {
                    if (i % j == 0) {
                        flag = true;
                        break;
                    }
                }
                if (flag)
                    comp++;
                else
                    prime++;
            }

            // Return the difference
            return comp - prime;
        }

        // Driver code
        let L = 2, R = 10;

        // Function call
        console.log(findDiff(L, R));

        // This code is contributed by rakeshsahni.

Output

1

Time Complexity: O((R-L) * sqrt( R))
Auxiliary Space: O(1)

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