# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to count the ` `// number of triplets in the ` `// array with GCD equal to K ` `#include ` `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; ` `} `

## Java

 `// 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 `

## Python3

 `# 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 `

Output:

```Number of triplets with GCD 3 are 4
```

Time Complexity: O (N * log N)
Auxiliary Space: O(N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : SHIVAMSINGH67, chitranayal