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Count the number of subsequences of length k having equal LCM and HCF
• Last Updated : 10 Oct, 2019

Given an array Arr and an integer K. The task is to find the number of subsequences of size K such that the LCM and HCF of the sequence is same.

Examples:

Input: Arr = {1, 2, 2, 3, 3}, K = 2
Output: 2

Subsequences are – {2, 2} and {3, 3}

Input: Arr = {1, 1, 1, 1, 2, 2}, K = 3
Output: 4

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

Approach:
LCM and HCF ( GCD ) are equal for a group of numbers when all the numbers are the same.
It can be proved in the following manner. Let’s take a case for two numbers.
Let the two numbers be x and y

HCF(x, y) = LCM(x, y) = k (say)

Since HCF(x, y) = k,
x = kn and, y = km, for some natural numbers m, n

We know, HCF × LCM = Product of the two numbers
Therefore, k2 = km × kn
Thus, mn = 1
Therefore, m = n = 1, since m, n are natural numbers
As a result,
x = kn = k, and, y = km = k
Implies, x = y = k, i.e. the numbers must be equal.

This concept can be extended to a group of numbers. Thus, it is proved that the numbers which are same have equal GCD and LCM.

After that, we need to find the frequencies of each of the elements in the array with the help of a map. Then, the formula of combinatorial theory will be used to find the count of subsequences of a particular length K for a particular element with a given frequency. This concept will be applied to all the elements with given frequencies and summation of all would be the answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation``#include ``using` `namespace` `std;`` ` `// Returns factorial of n``long` `long` `fact(``int` `n)``{``    ``long` `long` `res = 1;``    ``for` `(``int` `i = 2; i <= n; i++)``        ``res = res * i;``    ``return` `res;``}`` ` `// Returns nCr for the``// given values of r and n``long` `long` `nCr(``int` `n, ``int` `r)``{``    ``return` `fact(n) / (1LL * fact(r)``                      ``* fact(n - r));``}`` ` `long` `long` `number_of_subsequences(``int` `arr[],``                                 ``int` `k,``                                 ``int` `n)``{`` ` `    ``long` `long` `s = 0;`` ` `    ``// Map to store the frequencies``    ``// of each elements``    ``map<``int``, ``int``> m;`` ` `    ``// Loop to store the``    ``// frequencies of elements``    ``// in the map``    ``for` `(``int` `i = 0; i < n; i++) {``        ``m[arr[i]]++;``    ``}`` ` `    ``for` `(``auto` `j : m) {`` ` `        ``// Using nCR formula to``        ``// calculate the number``        ``// of subsequences of a``        ``// given length``        ``s = s + 1LL * nCr(j.second, k);``    ``}`` ` `    ``return` `s;``}`` ` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 1, 1, 1, 2, 2, 2 };``    ``int` `k = 2;``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);`` ` `    ``// Function calling``    ``cout << number_of_subsequences(arr, k, n);``    ``return` `0;``}`

## Java

 `// Java implementation for above approach``import` `java.util.*;``     ` `class` `GFG``{`` ` `// Returns factorial of n``static` `long` `fact(``int` `n)``{``    ``long` `res = ``1``;``    ``for` `(``int` `i = ``2``; i <= n; i++)``        ``res = res * i;``    ``return` `res;``}`` ` `// Returns nCr for the``// given values of r and n``static` `long` `nCr(``int` `n, ``int` `r)``{``    ``return` `fact(n) / (``1` `* fact(r) *``                          ``fact(n - r));``}`` ` `static` `long` `number_of_subsequences(``int` `arr[], ``                                   ``int` `k, ``int` `n)``{``    ``long` `s = ``0``;`` ` `    ``// Map to store the frequencies``    ``// of each elements``    ``HashMap mp = ``new` `HashMap();`` ` `    ``// Loop to store the``    ``// frequencies of elements``    ``// in the map``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``if``(mp.containsKey(arr[i]))``        ``{``            ``mp.put(arr[i], mp.get(arr[i]) + ``1``);``        ``}``        ``else``        ``{``            ``mp.put(arr[i], ``1``);``        ``}``    ``}`` ` `    ``for` `(Map.Entry j : mp.entrySet())``    ``{`` ` `        ``// Using nCR formula to``        ``// calculate the number``        ``// of subsequences of a``        ``// given length``        ``s = s + ``1` `* nCr(j.getValue(), k);``    ``}`` ` `    ``return` `s;``}`` ` `// Driver Code``static` `public` `void` `main ( String []arg)``{``    ``int` `arr[] = { ``1``, ``1``, ``1``, ``1``, ``2``, ``2``, ``2` `};``    ``int` `k = ``2``;``    ``int` `n = arr.length;`` ` `    ``// Function calling``    ``System.out.println(number_of_subsequences(arr, k, n));``}``}`` ` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of above approach`` ` `# Returns factorial of n``def` `fact(n):``    ``res ``=` `1``    ``for` `i ``in` `range``(``2``, n ``+` `1``):``        ``res ``=` `res ``*` `i``    ``return` `res`` ` `# Returns nCr for the``# given values of r and n``def` `nCr(n, r):``    ``return` `fact(n) ``/``/` `(fact(r) ``*` `fact(n ``-` `r))`` ` `def` `number_of_subsequences(arr, k, n):`` ` `    ``s ``=` `0`` ` `    ``# Map to store the frequencies``    ``# of each elements``    ``m ``=` `dict``()``  ` `    ``# Loop to store the``    ``# frequencies of elements``    ``# in the map``    ``for` `i ``in` `arr:``        ``m[i] ``=` `m.get(i, ``0``) ``+` `1`` ` `    ``for` `j ``in` `m:`` ` `        ``# Using nCR formula to``        ``# calculate the number``        ``# of subsequences of a``        ``# given length``        ``s ``=` `s ``+` `nCr(m[j], k)`` ` `    ``return` `s`` ` `# Driver Code``arr ``=` `[``1``, ``1``, ``1``, ``1``, ``2``, ``2``, ``2``]``k ``=` `2``n ``=` `len``(arr)`` ` `# Function calling``print``(number_of_subsequences(arr, k, n))`` ` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation for above approach``using` `System;``using` `System.Collections.Generic;``     ` `public` `class` `GFG``{``  ` `// Returns factorial of n``static` `long` `fact(``int` `n)``{``    ``long` `res = 1;``    ``for` `(``int` `i = 2; i <= n; i++)``        ``res = res * i;``    ``return` `res;``}``  ` `// Returns nCr for the``// given values of r and n``static` `long` `nCr(``int` `n, ``int` `r)``{``    ``return` `fact(n) / (1 * fact(r) *``                          ``fact(n - r));``}``  ` `static` `long` `number_of_subsequences(``int` `[]arr, ``                                   ``int` `k, ``int` `n)``{``    ``long` `s = 0;``  ` `    ``// Map to store the frequencies``    ``// of each elements``    ``Dictionary<``int``,``int``> mp = ``new` `Dictionary<``int``,``int``>();``    ``// Loop to store the``    ``// frequencies of elements``    ``// in the map``    ``for` `(``int` `i = 0 ; i < n; i++)``    ``{``        ``if``(mp.ContainsKey(arr[i]))``        ``{``            ``var` `val = mp[arr[i]];``            ``mp.Remove(arr[i]);``            ``mp.Add(arr[i], val + 1); ``        ``}``        ``else``        ``{``            ``mp.Add(arr[i], 1);``        ``}``    ``}``  ` `    ``foreach``(KeyValuePair<``int``, ``int``> j ``in` `mp)``    ``{``  ` `        ``// Using nCR formula to``        ``// calculate the number``        ``// of subsequences of a``        ``// given length``        ``s = s + 1 * nCr(j.Value, k);``    ``}``  ` `    ``return` `s;``}``  ` `// Driver Code``static` `public` `void` `Main ( String []arg)``{``    ``int` `[]arr = { 1, 1, 1, 1, 2, 2, 2 };``    ``int` `k = 2;``    ``int` `n = arr.Length;``  ` `    ``// Function calling``    ``Console.Write(number_of_subsequences(arr, k, n));``}``}``// This code is contributed by 29AjayKumar`
Output:
```9
```

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