# 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

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

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++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  ` `  ` `  `// 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; ` `} `

## Java

 `// 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. `

## Python 3

 `# Python 3 implementation of the approach ` `from` `math ``import` `gcd ` ` `  `# Function to return the count ` `# of quadruplets having gcd = k ` `def` `countQuadruplets(l, r, k): ` `     `  `    ``# To store the required count ` `    ``count ``=` `0` ` `  `    ``# Check every quadruplet pair ` `    ``# whether its gcd is k ` `    ``for` `u ``in` `range``(l, r ``+` `1` `,``1``): ` `        ``for` `v ``in` `range``(l, r ``+` `1``, ``1``): ` `            ``for` `w ``in` `range``(l, r ``+` `1``, ``1``): ` `                ``for` `x ``in` `range``(l, r ``+` `1``, ``1``): ` `                    ``if` `(gcd(gcd(u, v), gcd(w, x)) ``=``=` `k): ` `                        ``count ``+``=` `1` ` `  `    ``# Return the required count ` `    ``return` `count ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``l ``=` `1` `    ``r ``=` `10` `    ``k ``=` `2` ` `  `    ``print``(countQuadruplets(l, r, k)) ` `     `  `# This code is contributed  ` `# by Surendra_Gangwar `

## C#

 `// 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. `

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++

 `// C++ implementation of the approach ` `#include ` `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; ` `} `

## Java

 `// 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 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return ` `# the gcd of a and b ` `def` `gcd(a, b): ` `    ``if` `(b ``=``=` `0``): ` `        ``return` `a; ` `    ``return` `gcd(b, a ``%` `b); ` ` `  `# Function to return the count ` `# of quadruplets having gcd = k ` `def` `countQuadruplets(l, r, k): ` `    ``frequency ``=` `[``0``] ``*` `(r ``+` `1``); ` ` `  `    ``# Count the frequency of every possible gcd ` `    ``# value in the range ` `    ``for` `i ``in` `range``(l, r ``+` `1``): ` `        ``for` `j ``in` `range``(l, r ``+` `1``): ` `            ``frequency[gcd(i, j)] ``+``=` `1``; ` ` `  `    ``# To store the required count ` `    ``answer ``=` `0``; ` ` `  `    ``# Calculate the answer using frequency values ` `    ``for` `i ``in` `range``(l, r ``+` `1``): ` `        ``for` `j ``in` `range``(l, r ``+` `1``): ` `            ``if` `(gcd(i, j) ``=``=` `k): ` `                ``answer ``+``=` `(frequency[i] ``*` `frequency[j]); ` ` `  `    ``# Return the required count ` `    ``return` `answer; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``l, r, k ``=` `1``, ``10``, ``2``; ` ` `  `    ``print``(countQuadruplets(l, r, k)); ` ` `  `# This code is contributed by Rajput-Ji `

## C#

 `// 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 `

Output:

```607
```

Time Complexity: O((r – l)2) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.