Count of quadruplets from range [L, R] having GCD equal to K

Given an integer K and a range [L, R], the task is to count the quadruplet pairs from the given range having gcd equal to K.

Examples:

Input: L = 1, R = 5, K = 3
Output: 1
(3, 3, 3, 3) is the only valid quadruplet with gcd = 3



Input: L = 2, R = 24, K = 5
Output: 239

Naive approach: We can iterate over all the numbers with four loops and for every quadruplet pair check whether its gcd is equal to K.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Function to return
// the gcd of a and b
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
  
// Function to return the count
// of quadruplets having gcd = k
int countQuadruplets(int l, int r, int k)
{
  
    // To store the required count
    int count = 0;
  
    // Check every quadruplet pair
    // whether its gcd is k
    for (int u = l; u <= r; u++) {
        for (int v = l; v <= r; v++) {
            for (int w = l; w <= r; w++) {
                for (int x = l; x <= r; x++) {
                    if (gcd(gcd(u, v), gcd(w, x)) == k)
                        count++;
                }
            }
        }
    }
  
    // Return the required count
    return count;
}
  
// Driver code
int main()
{
    int l = 1, r = 10, k = 2;
  
    cout << countQuadruplets(l, r, k);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.io.*;
  
class GFG {
  
    // Function to return
    // the gcd of a and b
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
  
    // Function to return the count
    // of quadruplets having gcd = k
    static int countQuadruplets(int l, int r, int k)
    {
  
        // To store the required count
        int count = 0;
  
        // Check every quadruplet pair
        // whether its gcd is k
        for (int u = l; u <= r; u++) {
            for (int v = l; v <= r; v++) {
                for (int w = l; w <= r; w++) {
                    for (int x = l; x <= r; x++) {
                        if (gcd(gcd(u, v), gcd(w, x)) == k)
                            count++;
                    }
                }
            }
        }
  
        // Return the required count
        return count;
    }
  
    // Driver code
    public static void main(String[] args)
    {
  
        int l = 1, r = 10, k = 2;
  
        System.out.println(countQuadruplets(l, r, k));
    }
}
  
// This code is contributed by jit_t.

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C#

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// C# implementation of the approach
using System;
  
class GFG {
  
    // Function to return
    // the gcd of a and b
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
  
    // Function to return the count
    // of quadruplets having gcd = k
    static int countQuadruplets(int l, int r, int k)
    {
  
        // To store the required count
        int count = 0;
  
        // Check every quadruplet pair
        // whether its gcd is k
        for (int u = l; u <= r; u++) {
            for (int v = l; v <= r; v++) {
                for (int w = l; w <= r; w++) {
                    for (int x = l; x <= r; x++) {
                        if (gcd(gcd(u, v), gcd(w, x)) == k)
                            count++;
                    }
                }
            }
        }
  
        // Return the required count
        return count;
    }
  
    // Driver code
    static public void Main()
    {
        int l = 1, r = 10, k = 2;
        Console.WriteLine(countQuadruplets(l, r, k));
    }
}
  
// This code is contributed by ajit.

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Output:

607

Time Complexity: O((r – l)4)

Efficient approach:

  1. Find the GCD of every possible pair (x, y) in the given range.
  2. Count the frequencies of every possible GCD value.
  3. After that if the GCD value of two numbers is k then increment count by frequency[i] * frequency[j].

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Function to return
// the gcd of a and b
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
  
// Function to return the count
// of quadruplets having gcd = k
int countQuadruplets(int l, int r, int k)
{
  
    int frequency[r + 1] = { 0 };
  
    // Count the frequency of every possible gcd
    // value in the range
    for (int i = l; i <= r; i++) {
        for (int j = l; j <= r; j++) {
            frequency[gcd(i, j)]++;
        }
    }
  
    // To store the required count
    long long answer = 0;
  
    // Calculate the answer using frequency values
    for (int i = 1; i <= r; i++) {
        for (int j = 1; j <= r; j++) {
            if (gcd(i, j) == k) {
                answer += (frequency[i] * frequency[j]);
            }
        }
    }
  
    // Return the required count
    return answer;
}
  
// Driver code
int main()
{
    int l = 1, r = 10, k = 2;
  
    cout << countQuadruplets(l, r, k);
  
    return 0;
}

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Java

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// Java implementation of the approach 
class GFG
{
      
// Function to return 
// the gcd of a and b 
static int gcd(int a, int b) 
    if (b == 0
        return a; 
    return gcd(b, a % b); 
  
// Function to return the count 
// of quadruplets having gcd = k 
static int countQuadruplets(int l, int r, int k) 
  
    int frequency[]= new int[r + 1] ; 
  
    // Count the frequency of every possible gcd 
    // value in the range 
    for (int i = l; i <= r; i++)
    
        for (int j = l; j <= r; j++) 
        
            frequency[gcd(i, j)]++; 
        
    
  
    // To store the required count 
    long answer = 0
  
    // Calculate the answer using frequency values 
    for (int i = 1; i <= r; i++)
    
        for (int j = 1; j <= r; j++) 
        
            if (gcd(i, j) == k) 
            
                answer += (frequency[i] * frequency[j]); 
            
        
    
  
    // Return the required count 
    return (int)answer; 
  
// Driver code 
public static void main(String args[])
    int l = 1, r = 10, k = 2
  
    System.out.println(countQuadruplets(l, r, k)); 
}
  
// This code is contributed by Arnab Kundu

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C#

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// C# implementation of the approach 
using System;
  
class GFG
{
          
// Function to return 
// the gcd of a and b 
static int gcd(int a, int b) 
    if (b == 0) 
        return a; 
    return gcd(b, a % b); 
  
// Function to return the count 
// of quadruplets having gcd = k 
static int countQuadruplets(int l, int r, int k) 
  
    int []frequency= new int[r + 1] ; 
  
    // Count the frequency of every possible gcd 
    // value in the range 
    for (int i = l; i <= r; i++)
    
        for (int j = l; j <= r; j++) 
        
            frequency[gcd(i, j)]++; 
        
    
  
    // To store the required count 
    long answer = 0; 
  
    // Calculate the answer using frequency values 
    for (int i = 1; i <= r; i++)
    
        for (int j = 1; j <= r; j++) 
        
            if (gcd(i, j) == k) 
            
                answer += (frequency[i] * frequency[j]); 
            
        
    
  
    // Return the required count 
    return (int)answer; 
  
// Driver code 
static public void Main ()
{
    int l = 1, r = 10, k = 2; 
    Console.WriteLine(countQuadruplets(l, r, k)); 
}
  
// This code is contributed by @ajit_00023

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Output:

607

Time Complexity: O((r – l)2)



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