# Count number of subsets of a set with GCD equal to a given number

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}

**We strongly recommend you to minimize your browser and try this yourself first.**

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<bits/stdc++.h>` `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<` `int` `, ` `int` `> freq;` ` ` `// Map to store number of subsets with given gcd` ` ` `unordered_map<` `int` `, ` `int` `> 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<n; i++)` ` ` `{` ` ` `arrMax = max(arrMax, arr[i]);` ` ` `freq[arr[i]]++;` ` ` `}` ` ` `// 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 = 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} considering i = 2 and` ` ` `// subset we need to exclude are those having gcd as 4` ` ` `sub += subsets[j*i];` ` ` `}` ` ` ` ` `// Number of subsets with GCD equal to 'i' is pow(2, add)` ` ` `// - 1 - sub ` ` ` `subsets[i] = (1<<add) - 1 - sub;` ` ` `}` ` ` `for` `(` `int` `i=0; i<m; i++)` ` ` `cout << ` `"Number of subsets with gcd "` `<< gcd[i] << ` `" is "` ` ` `<< subsets[gcd[i]] << endl;` `}` `// Driver program` `int` `main()` `{` ` ` `int` `gcd[] = {2, 3};` ` ` `int` `arr[] = {9, 6, 2};` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `int` `m = ` `sizeof` `(gcd)/` `sizeof` `(gcd[0]);` ` ` `ccountSubsets(arr, n, gcd, m);` ` ` `return` `0;` `}` |

## 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<Integer,` ` ` `Integer> freq =` ` ` `new` `HashMap<Integer,` ` ` `Integer>();` ` ` `// Map to store number of` ` ` `// subsets with given gcd` ` ` `HashMap<Integer,` ` ` `Integer> subsets =` ` ` `new` `HashMap<Integer,` ` ` `Integer>();` ` ` `// 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}` ` ` `// considering i = 2 and` ` ` `// subset we need to exclude` ` ` `// are those having 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} considering i = 2 and` ` ` `# subset we need to exclude are those` ` ` `# having 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<` `int` `,` ` ` `int` `> freq =` ` ` `new` `Dictionary<` `int` `,` ` ` `int` `>();` ` ` `// Map to store number of` ` ` `// subsets with given gcd` ` ` `Dictionary<` `int` `,` ` ` `int` `> subsets =` ` ` `new` `Dictionary<` `int` `,` ` ` `int` `>();` ` ` `// 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}` ` ` `// considering i = 2 and` ` ` `// subset we need to exclude` ` ` `// are those having 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` |

## Javascript

`<script>` `// JavaScript program to count number of` `// subsets with given GCDs` `// n is size of arr[] and m is sizeof gcd[]` `function` `countSubsets(arr, n, gcd, m){` ` ` `// Map to store frequency of array elements` ` ` `let freq = ` `new` `Map` ` ` `// Map to store number of subsets` ` ` `// with given gcd` ` ` `let subsets = ` `new` `Map` ` ` `// Initialize maximum element.` ` ` `// Assumption: all array elements` ` ` `// are positive.` ` ` `let arrMax = 0` ` ` `// Find maximum element in array and` ` ` `// fill frequency map.` ` ` `for` `(let i = 0; i < n; i++)` ` ` `{` ` ` `arrMax = Math.max(arrMax, arr[i])` ` ` `if` `(freq.has(arr[i]) == ` `false` `)` ` ` `freq.set(arr[i], 1)` ` ` `else` ` ` `freq.set(arr[i], freq.get(arr[i])+1)` ` ` `}` ` ` `// Run a loop from max element to 1` ` ` `// to find subsets with all gcds` ` ` `for` `(let i = arrMax; i > 0; i--)` ` ` `{` ` ` `let sub = 0` ` ` `let add = 0` ` ` `if` `(freq.has(i))` ` ` `add = freq.get(i)` ` ` `let 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` `(freq.has(j * i))` ` ` `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} considering i = 2 and` ` ` `// subset we need to exclude are those` ` ` `// having gcd as 4` ` ` `sub += subsets.get(j * i)` ` ` `j += 1` ` ` `}` ` ` `// Number of subsets with GCD equal` ` ` `// to 'i' is pow(2, add) - 1 - sub` ` ` `subsets.set(i , (1 << add) - 1 - sub)` ` ` `}` ` ` `for` `(let i = 0; i < m; i++)` ` ` `{` ` ` `document.write( `Number of subsets ` `with` `gcd ${gcd[i]} is ${subsets.get(gcd[i])}`,` `"</br>"` `)` ` ` `}` `}` `// Driver Code` `let gcd = [2, 3]` `let arr = [9, 6, 2]` `let n = arr.length` `let m = gcd.length` `countSubsets(arr, n, gcd, m)` `// This code is contributed by shinjanpatra` `</script>` |

**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.

This article is contributed by **Ekta Goel**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.