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Count number of subsets of a set with GCD equal to a given number
• Difficulty Level : Hard
• Last Updated : 06 Nov, 2020

Given a set of positive integer elements, find the count of subsets with GCDs equal to given numbers.
Examples:

Input:  arr[] = {2, 3, 4}, gcd[] = {2, 3}
Output: Number of subsets with gcd 2 is 2
Number of subsets with gcd 3 is 1
The two subsets with GCD equal to 2 are {2} and {2, 4}.
The one subset with GCD equal to 3 ss {3}.

Input:  arr[] = {6, 3, 9, 2}, gcd = {3, 2}
Output: Number of subsets with gcd 3 is 5
Number of subsets with gcd 2 is 2
The five subsets with GCD equal to 3 are {3}, {6, 3},
{3, 9}, {6, 9) and {6, 3, 9}.
The two subsets with GCD equal to 2 are {2} and {2, 6}

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A Simple Solution is to generate all subsets of given set and find GCD of every subset.
Below is an Efficient Solution for small numbers, i,e, the maximum of all numbers is not very high.

1) Find the maximum number of given numbers.  Let the
maximum be arrMax.
2) Count occurrences of all numbers using a hash. Let
this hash be 'freq'
3) The maximum possible GCD can be arrMax. Run a loop
for i = arrMax to 1
a) Count number of subsets for current GCD.
4) Now we have counts for all possible GCDs, return
count for given gcds.

How does step 3.a work?
How to get the number of subsets for a given GCD ‘i’ where i lies in the range from 1 to arrMax. The idea is to count all multiples of i using ‘freq’ built-in step 2. Let there be ‘add’ multiples of i. The number of all possible subsets with ‘add’ numbers would be “pow(2, add) – 1”, excluding the empty set. For example, if given array is {2, 3, 6} and i = 3, there are 2 multiples of 3 (3 and 6). So there will be 3 subsets {3}, {3, 6} and {6} which have a multiple of i as GCD. These subsets also include {6} which doesn’t have 3 as GCD, but a multiple of 3. So we need to subtract such subsets. We store subset counts for every GCD in another hash map ‘subset’. Let ‘sub’ be the number of subsets that have multiple of ‘i’ as GCD. The value of ‘sub’ for any multiple of ‘i’ can directly be obtained from subset[] as we are evaluating counts from arrMax to 1.
Below is the implementation of the above idea.

## C++

 // C++ program to count number of subsets with given GCDs #include using namespace std;   // n is size of arr[] and m is sizeof gcd[] void ccountSubsets(int arr[], int n, int gcd[], int m) {     // Map to store frequency of array elements     unordered_map freq;       // Map to store number of subsets with given gcd     unordered_map subsets;       // Initialize maximum element. Assumption: all array     // elements are positive.     int arrMax = 0;       // Find maximum element in array and fill frequency     // map.     for (int i=0; i=1; i--)     {         int sub = 0;         int add = freq[i];           // Run a loop for all multiples of i         for (int j = 2; j*i <= arrMax; j++)         {             // Sum the frequencies of every element which             // is a multiple of i             add += freq[j*i];               // Excluding those subsets which have gcd > i but             // not i i.e. which have gcd as multiple of i in             // the subset for ex: {2,3,4} cnsidering i = 2 and             // subset we need to exclude are those havng gcd as 4             sub += subsets[j*i];         }                   // Number of subsets with GCD equal to 'i' is pow(2, add)         // - 1 - sub            subsets[i] = (1<

## Java

 // Java program to count // number of subsets with // given GCDs import java.util.*; class GFG{   // n is size of arr[] and // m is sizeof gcd[] static void ccountSubsets(int arr[], int n,                           int gcd[], int m) {   // Map to store frequency   // of array elements   HashMap freq =           new HashMap();     // Map to store number of   // subsets with given gcd   HashMap subsets =           new HashMap();     // Initialize maximum element.   // Assumption: all array   // elements are positive.   int arrMax = 0;     // Find maximum element in   // array and fill frequency   // map.   for (int i = 0; i < n; i++)   {     arrMax = Math.max(arrMax,                       arr[i]);     if(freq.containsKey(arr[i]))     {       freq.put(arr[i],       freq.get(arr[i]) + 1);     }     else     {       freq.put(arr[i], 1);     }   }     // Run a loop from max element   // to 1 to find subsets   // with all gcds   for (int i = arrMax; i >= 1; i--)   {     int sub = 0;     int add = 0;     if(freq.containsKey(i))       add = freq.get(i);       // Run a loop for all multiples     // of i     for (int j = 2; j * i <= arrMax; j++)     {       // Sum the frequencies of       // every element which       // is a multiple of i       if(freq.containsKey(i * j))         add += freq.get(j * i);         // Excluding those subsets       // which have gcd > i but       // not i i.e. which have       // gcd as multiple of i in       // the subset for ex: {2,3,4}       // cnsidering i = 2 and       // subset we need to exclude       // are those havng gcd as 4       sub += subsets.get(j * i);     }       // Number of subsets with GCD     // equal to 'i' is Math.pow(2, add)     // - 1 - sub       subsets.put(i, (1 << add) -                 1 - sub);   }     for (int i = 0; i < m; i++)     System.out.print("Number of subsets with gcd " +                       gcd[i] + " is " +                      subsets.get(gcd[i]) + "\n"); }   // Driver program public static void main(String[] args) {   int gcd[] = {2, 3};   int arr[] = {9, 6, 2};   int n = arr.length;   int m = gcd.length;   ccountSubsets(arr, n, gcd, m); } }   // This code is contributed by shikhasingrajput

## Python3

 # Python3 program to count number of # subsets with given GCDs   # n is size of arr[] and m is sizeof gcd[] def countSubsets(arr, n, gcd, m):       # Map to store frequency of array elements     freq = dict()       # Map to store number of subsets     # with given gcd     subsets = dict()       # Initialize maximum element.     # Assumption: all array elements     # are positive.     arrMax = 0       # Find maximum element in array and     # fill frequency map.     for i in range(n):         arrMax = max(arrMax, arr[i])         if arr[i] not in freq:             freq[arr[i]] = 1         else:             freq[arr[i]] += 1       # Run a loop from max element to 1     # to find subsets with all gcds     for i in range(arrMax, 0, -1):         sub = 0         add = 0         if i in freq:             add = freq[i]         j = 2           # Run a loop for all multiples of i         while j * i <= arrMax:               # Sum the frequencies of every element             # which is a multiple of i             if j * i in freq:                 add += freq[j * i]               # Excluding those subsets which have             # gcd > i but not i i.e. which have gcd             # as multiple of i in the subset.             # for ex: {2,3,4} cnsidering i = 2 and             # subset we need to exclude are those             # havng gcd as 4             sub += subsets[j * i]             j += 1           # Number of subsets with GCD equal         # to 'i' is pow(2, add) - 1 - sub         subsets[i] = (1 << add) - 1 - sub       for i in range(m):         print("Number of subsets with gcd %d is %d" %              (gcd[i], subsets[gcd[i]]))   # Driver Code if __name__ == "__main__":     gcd = [2, 3]     arr = [9, 6, 2]     n = len(arr)     m = len(gcd)     countSubsets(arr, n, gcd, m)   # This code is contributed by # sanjeev2552

## C#

 // C# program to count // number of subsets with // given GCDs using System; using System.Collections.Generic; class GFG{   // n is size of []arr and // m is sizeof gcd[] static void ccountSubsets(int []arr, int n,                           int []gcd, int m) {   // Map to store frequency   // of array elements   Dictionary freq =              new Dictionary();     // Map to store number of   // subsets with given gcd   Dictionary subsets =              new Dictionary();     // Initialize maximum element.   // Assumption: all array   // elements are positive.   int arrMax = 0;     // Find maximum element in   // array and fill frequency   // map.   for (int i = 0; i < n; i++)   {     arrMax = Math.Max(arrMax,                       arr[i]);     if(freq.ContainsKey(arr[i]))     {       freq.Add(arr[i],       freq[arr[i]] + 1);     }     else     {       freq.Add(arr[i], 1);     }   }     // Run a loop from max element   // to 1 to find subsets   // with all gcds   for (int i = arrMax; i >= 1; i--)   {     int sub = 0;     int add = 0;     if(freq.ContainsKey(i))       add = freq[i];       // Run a loop for all     // multiples of i     for (int j = 2;              j * i <= arrMax; j++)     {       // Sum the frequencies of       // every element which       // is a multiple of i       if(freq.ContainsKey(i * j))         add += freq[j * i];         // Excluding those subsets       // which have gcd > i but       // not i i.e. which have       // gcd as multiple of i in       // the subset for ex: {2,3,4}       // cnsidering i = 2 and       // subset we need to exclude       // are those havng gcd as 4       sub += subsets[j * i];     }       // Number of subsets with GCD     // equal to 'i' is Math.Pow(2, add)     // - 1 - sub       subsets.Add(i, (1 << add) -                 1 - sub);   }     for (int i = 0; i < m; i++)     Console.Write("Number of subsets with gcd " +                     gcd[i] + " is " +                    subsets[gcd[i]] + "\n"); }   // Driver code public static void Main(String[] args) {   int []gcd = {2, 3};   int []arr = {9, 6, 2};   int n = arr.Length;   int m = gcd.Length;   ccountSubsets(arr, n, gcd, m); } }   // This code is contributed by Rajput-Ji

Output:

Number of subsets with gcd 2 is 2
Number of subsets with gcd 3 is 1

Exercise: Extend the above solution so that all calculations are done under modulo 1000000007 to avoid overflows.