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<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 programint 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 GCDsimport 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 programpublic 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 Codeif __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 GCDsusing 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 codepublic 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 Codelet gcd = [2, 3]let arr = [9, 6, 2]let n = arr.lengthlet m = gcd.lengthcountSubsets(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
Time Complexity: O(n * log(max(arr)) * sqrt(max(arr))), where n is the size of the input array and max(arr) represents the maximum element in the array.
Auxiliary Space: O(max(arr)), as it uses two unordered maps to store frequency of array elements and the number of subsets with a given gcd. The space required by the maps depends on the maximum element in the array.
Exercise: Extend the above solution so that all calculations are done under modulo 1000000007 to avoid overflows.
This article is contributed by Aarti_Rathi and Ekta Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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