Count of triplets in a given Array having GCD K

Given an integer array arr[] and an integer K, the task is to count all triplets whose GCD is equal to K.

Examples:

Input: arr[] = {1, 4, 8, 14, 20}, K = 2
Output: 3
Explanation:
Triplets (4, 14, 20), (8, 14, 20) and (4, 8, 14) have GCD equal to 2.

Input: arr[] = {1, 2, 3, 4, 5}, K = 7
Output: 0

Approach:
Follow the steps below to solve the problem:



  1. Maintain an array cnt[i] which denotes the numbers of triplet in array with GCD = i.
  2. Now, to find cnt[i], first of all count all the multiples of i in the array which is stored in another array mul[i].
  3. Now select any three values from mul[i]. Number of ways of selecting three values is mul [i] C 3, store this value in cnt[i].
  4. But it also includes the triplets with GCD a multiple of i.So we Again iterate over all the multiples of i and subtract cnt[j] from cnt[i] where j is a multiple of i.
  5. Finaly return cnt[K].

Below is the implementation of the above approach:

C++

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// C++ program to count the
// number of triplets in the
// array with GCD equal to K
#include <bits/stdc++.h>
using namespace std;
  
const int MAXN = 1e6 + 1;
  
// frequency array
int freq[MAXN] = { 0 };
  
// mul[i] stores the count
// of multiples of i
int mul[MAXN] = { 0 };
  
// cnt[i] stores the count
// of triplets with gcd = i
int cnt[MAXN] = { 0 };
  
// Return nC3
int nC3(int n)
{
    if (n < 3)
        return 0;
    return (n * (n - 1) * (n - 2)) / 6;
}
  
// Function to count and return
// the number of triplets in the
// array with GCD equal to K
void count_triplet(vector<int> arr,
                   int N, int K)
{
    for (int i = 0; i < N; i++) {
  
        // Store frequency of
        // array elements
        freq[arr[i]]++;
    }
    for (int i = 1; i <= 1000000; i++) {
        for (int j = i; j <= 1000000;
             j += i) {
            // Store the multiples of
            // i present in the array
            mul[i] += freq[j];
        }
        // Count triplets with gcd
        // equal to any multiple of i
        cnt[i] = nC3(mul[i]);
    }
  
    // Remove all triplets which have gcd
    // equal to a multiple of i
    for (int i = 1000000; i >= 1; i--) {
        for (int j = 2 * i; j <= 1000000;
             j += i) {
            cnt[i] -= cnt[j];
        }
    }
    cout << "Number of triplets "
         << "with GCD " << K;
    cout << " are " << cnt[K];
}
// Driver Program
int main()
{
    vector<int> arr = { 1, 7, 12, 6,
                        15, 9 };
    int N = 6, K = 3;
    count_triplet(arr, N, K);
    return 0;
}

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Java

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// Java program to count the
// number of triplets in the
// array with GCD equal to K
class GFG{
      
static int MAXN = 1000001;
  
// Frequency array
static int freq[] = new int[MAXN];
  
// mul[i] stores the count
// of multiples of i
static int mul[] = new int[MAXN];
  
// cnt[i] stores the count
// of triplets with gcd = i
static int cnt[] = new int[MAXN];
  
// Return nC3
static int nC3(int n)
{
    if (n < 3)
        return 0;
          
    return (n * (n - 1) * (n - 2)) / 6;
}
  
// Function to count and return
// the number of triplets in the
// array with GCD equal to K
static void count_triplet(int[] arr,
                          int N, int K)
    int i, j;
    for(i = 0; i < N; i++)
    {
         
       // Store frequency of
       // array elements
       freq[arr[i]]++;
    }
      
    for(i = 1; i <= 1000000; i++)
    {
       for(j = i; j <= 1000000; j += i)
       {
             
          // Store the multiples of
          // i present in the array
          mul[i] += freq[j];
       }
         
       // Count triplets with gcd
       // equal to any multiple of i
       cnt[i] = nC3(mul[i]);
    }
  
    // Remove all triplets which have gcd
    // equal to a multiple of i
    for(i = 1000000; i >= 1; i--)
    {
       for(j = 2 * i; j <= 1000000; j += i)
       {
          cnt[i] -= cnt[j];
       }
    }
    System.out.print("Number of triplets "
                     "with GCD " + K);
    System.out.print(" are " + cnt[K]);
}
  
// Driver code
public static void main (String []args)
{
    int []arr = { 1, 7, 12, 6, 15, 9 };
    int N = 6, K = 3;
      
    count_triplet(arr, N, K);
}
  
// This code is contributed by chitranayal

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Python3

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# Python3 program to count the number of 
# triplets in the array with GCD equal to K 
MAXN = int(1e6 + 1)
  
# Frequency array
freq = [0] * MAXN
  
# mul[i] stores the count
# of multiples of i
mul = [0] * MAXN
  
# cnt[i] stores the count
# of triplets with gcd = i
cnt = [0] * MAXN
  
# Return nC3
def nC3(n):
  
    if(n < 3):
        return 0
    return (n * (n - 1) * (n - 2)) / 6
  
# Function to count and return
# the number of triplets in the
# array with GCD equal to K
def count_triplet(arr, N, K):
  
    for i in range(N):
  
        # Store frequency of
        # array elements
        freq[arr[i]] += 1
  
    for i in range(1, 1000000 + 1):
        for j in range(i, 1000000 + 1, i):
  
            # Store the multiples of
            # i present in the array
            mul[i] += freq[j]
  
        # Count triplets with gcd
        # equal to any multiple of i
        cnt[i] = nC3(mul[i])
  
    # Remove all triplets which have gcd
    # equal to a multiple of i
    for i in range(1000000, 0, -1):
        for j in range(2 * i, 1000000 + 1, i):
            cnt[i] -= cnt[j]
  
    print("Number of triplets with GCD"
          " {0} are {1}".format(K, int(cnt[K])))
  
# Driver Code
if __name__ == '__main__':
  
    arr = [ 1, 7, 12, 6, 15, 9 ]
    N = 6
    K = 3
  
    count_triplet(arr, N, K)
  
# This code is contributed by Shivam Singh

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Output:

Number of triplets with GCD 3 are 4

Time Complexity: O (N * log N)
Auxiliary Space: O(N)

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Improved By : SHIVAMSINGH67, chitranayal