Queries to count Composite Magic Numbers from a given range [L, R]

Last Updated : 07 Mar, 2022

Given two arrays L[] and R[] of sizes Q, the task is to find the number of composite magic numbers i.e numbers which are both composite numbers as well as Magic numbers from the range [L[i], R[i]] ( 0 ? i < Q).

Examples:

Input: Q = 1, L[] = {10}, R[] = {100}
Output: 8
Explanation: Numbers in the range [L[0], R[0]] which are both composite as well as Magic numbers are {10, 28, 46, 55, 64, 82, 91, 100}.

Input: Q = 1, L[] = {1200}, R[] = {1300}
Output: 9
Explanation: Numbers in the range [L[0], R[0]] which are both composite as well as Magic numbers are {1207, 1216, 1225, 1234, 1243, 1252, 1261, 1270, 1288}.

Naive Approach: The simplest approach to solve the problem is to traverse all the numbers in the range [L[i], R[i]] (0 ? i < Q) and for every number, check if it is composite as well as a magic number or not.

Time Complexity: O(Q * N^3/2)
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, precompute and store the counts of Composite Magic Numbers in an array over a certain range. This optimizes the computational complexities of each query to O(1). Follow the steps below to solve the problem:

Initialize an integer array dp[] such that dp[i] stores the count of Composite Magic Numbers up to i.
Traverse the range [1, 10⁶] and for every number in the range, check if it is a Composite Magic Number or not. If found to be true, update dp[i] with dp[i - 1] + 1 . Otherwise, update dp[i] with dp[i - 1]
Traverse the arrays L[] and R[] simultaneously and print dp[R[i]] - dp[L[i] - 1] as the required answer.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <iostream>
using namespace std;

// Check if a number is magic
// number or not
bool isMagic(int num)
{
    return (num % 9 == 1);
}

// Check number is composite
// number or not
bool isComposite(int n)
{
    // Corner cases
    if (n <= 1)
        return false;

    if (n <= 3)
        return false;

    // Check if the number is
    // a multiple of 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return true;

    // Check for multiples of remaining primes
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return true;

    return false;
}

// Function to find Composite Magic
// Numbers in given ranges
void find(int L[], int R[], int q)
{
    // dp[i]: Stores the count of
    // composite Magic numbers up to i
    int dp[1000005];

    dp[0] = 0;
    dp[1] = 0;

    // Traverse in the range [1, 1e5)
    for (int i = 1; i < 1000005; i++) {

        // Check if number is Composite number
        // as well as a Magic number or not
        if (isComposite(i) && isMagic(i)) {
            dp[i] = dp[i - 1] + 1;
        }

        else
            dp[i] = dp[i - 1];
    }

    // Print results for each query
    for (int i = 0; i < q; i++)
        cout << dp[R[i]] - dp[L[i] - 1] << endl;
}

// Driver Code
int main()
{
    int L[] = { 10, 3 };
    int R[] = { 100, 2279 };
    int Q = 2;

    find(L, R, Q);
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG
{
    
    // Check if a number is magic
    // number or not
    static boolean isMagic(int num)
    {
        return (num % 9 == 1);
    }
    
    // Check number is composite
    // number or not
    static boolean isComposite(int n)
    {
        // Corner cases
        if (n <= 1)
            return false;
    
        if (n <= 3)
            return false;
    
        // Check if the number is
        // a multiple of 2 or 3
        if (n % 2 == 0 || n % 3 == 0)
            return true;
    
        // Check for multiples of remaining primes
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return true;
    
        return false;
    }
    
    // Function to find Composite Magic
    // Numbers in given ranges
    static void find(int L[], int R[], int q)
    {
        // dp[i]: Stores the count of
        // composite Magic numbers up to i
        int dp[] = new int[1000005];
    
        dp[0] = 0;
        dp[1] = 0;
    
        // Traverse in the range [1, 1e5)
        for (int i = 1; i < 1000005; i++)
        {
    
            // Check if number is Composite number
            // as well as a Magic number or not
            if (isComposite(i) && isMagic(i) == true) 
            {
                dp[i] = dp[i - 1] + 1;
            }
    
            else
                dp[i] = dp[i - 1];
        }
    
        // Print results for each query
        for (int i = 0; i < q; i++)
            System.out.println(dp[R[i]] - dp[L[i] - 1]);
    }
    
    // Driver Code
    public static void main (String[] args)
    {
        int L[] = { 10, 3 };
        int R[] = { 100, 2279 };
        int Q = 2;
    
        find(L, R, Q);
    }
}

// This code is contributed by AnkThon

Python3

# Python3 program to implement
# the above approach

# Check if a number is magic
# number or not
def isMagic(num):
    
    return (num % 9 == 1)

# Check number is composite
# number or not
def isComposite(n):
    
    # Corner cases
    if (n <= 1):
        return False

    if (n <= 3):
        return False

    # Check if the number is
    # a multiple of 2 or 3
    if (n % 2 == 0 or n % 3 == 0):
        return True

    # Check for multiples of remaining primes
    for i in range(5, n + 1, 6):
        if i * i > n + 1:
            break
        
        if (n % i == 0 or n % (i + 2) == 0):
            return True

    return False

# Function to find Composite Magic
# Numbers in given ranges
def find(L, R, q):
    
    # dp[i]: Stores the count of
    # composite Magic numbers up to i
    dp = [0] * 1000005

    dp[0] = 0
    dp[1] = 0

    # Traverse in the range [1, 1e5)
    for i in range(1, 1000005):
        
        # Check if number is Composite number
        # as well as a Magic number or not
        if (isComposite(i) and isMagic(i)):
            dp[i] = dp[i - 1] + 1
        else:
            dp[i] = dp[i - 1]

    # Print results for each query
    for i in range(q):
        print(dp[R[i]] - dp[L[i] - 1])

# Driver Code
if __name__ == '__main__':
    
    L = [ 10, 3 ]
    R = [ 100, 2279 ]
    Q = 2

    find(L, R, Q)

# This code is contributed by mohit kumar 29

// C# program to implement
// the above approach  
using System;

class GFG{
     
// Check if a number is magic
// number or not
static bool isMagic(int num)
{
    return (num % 9 == 1);
}
 
// Check number is composite
// number or not
static bool isComposite(int n)
{
    
    // Corner cases
    if (n <= 1)
        return false;
 
    if (n <= 3)
        return false;
 
    // Check if the number is
    // a multiple of 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return true;
 
    // Check for multiples of remaining primes
    for(int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return true;
 
    return false;
}
 
// Function to find Composite Magic
// Numbers in given ranges
static void find(int[] L, int[] R, int q)
{
    
    // dp[i]: Stores the count of
    // composite Magic numbers up to i
    int[] dp = new int[1000005];
 
    dp[0] = 0;
    dp[1] = 0;
 
    // Traverse in the range [1, 1e5)
    for(int i = 1; i < 1000005; i++)
    {
        
        // Check if number is Composite number
        // as well as a Magic number or not
        if (isComposite(i) && isMagic(i) == true) 
        {
            dp[i] = dp[i - 1] + 1;
        }
 
        else
            dp[i] = dp[i - 1];
    }
 
    // Print results for each query
    for(int i = 0; i < q; i++)
        Console.WriteLine(dp[R[i]] - dp[L[i] - 1]);
}
 
// Driver Code
public static void Main ()
{
    int[] L = { 10, 3 };
    int[] R = { 100, 2279 };
    int Q = 2;
 
    find(L, R, Q);
}
}

// This code is contributed by susmitakundugoaldanga

JavaScript

<script>

// JavaScript program to implement
// the above approach
    
    // Check if a number is magic
    // number or not
    function isMagic(num)
    {
        return (num % 9 == 1);
    }
     
    // Check number is composite
    // number or not
    function isComposite(n)
    {
        // Corner cases
        if (n <= 1)
            return false;
     
        if (n <= 3)
            return false;
     
        // Check if the number is
        // a multiple of 2 or 3
        if (n % 2 == 0 || n % 3 == 0)
            return true;
     
        // Check for multiples of remaining primes
        for (let i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return true;
     
        return false;
    }
     
    // Function to find Composite Magic
    // Numbers in given ranges
    function find(L, R, q)
    {
        // dp[i]: Stores the count of
        // composite Magic numbers up to i
        let dp = [];
     
        dp[0] = 0;
        dp[1] = 0;
     
        // Traverse in the range [1, 1e5)
        for (let i = 1; i < 1000005; i++)
        {
     
            // Check if number is Composite number
            // as well as a Magic number or not
            if (isComposite(i) && isMagic(i) == true)
            {
                dp[i] = dp[i - 1] + 1;
            }
     
            else
                dp[i] = dp[i - 1];
        }
     
        // Print results for each query
        for (let i = 0; i < q; i++)
            document.write(dp[R[i]] - dp[L[i] - 1] + "<br/>");
    }
     

     
// Driver code
        
        let L = [ 10, 3 ];
        let R = [100, 2279 ];
        let Q = 2;
     
        find(L, R, Q);

</script>

Output:

8
198

Time Complexity: O(N^3/2)
Auxiliary Space: O(N)

Count of Double Prime numbers in a given range L to R

2018kucp1001

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Queries to count Composite Magic Numbers from a given range [L, R]

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