Find all possible GCDs of every subsequences of given Array
Given an array arr[] consisting of N positive integers, the task is to find all the possible distinct Greatest Common Divisors(GCDs) among all the non-empty subsequences of the array arr[].
Examples:
Input: arr[] = {3, 4, 8}
Output: 1 3 4 8
Explanation:
The non-empty subsequences possible are {3}, {4}, {8}, {3, 4}, {4, 8}, {3, 8}, {3, 4, 8} and their corresponding GCDs are 3, 4, 8, 1, 4, 1, 1.
Therefore, print all the GCDs as {1, 3, 4, 8}.Input: arr[] = {3, 8, 9, 4, 13, 45, 6}
Output: 1 2 3 4 6 8 9 13 45
Naive Approach: The simplest approach to solve the given problem is to generate all possible subsequences of the given array and store all the GCDs of the subsequence in a set. After checking for all the subsequences, print the element stores in the set as all possible GCDs that can be formed.
Time Complexity: O(log M*2N), where M is the maximum element of the array.
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized using Greedy Approach by observing the fact that the GCD of any subsequence lies over the range [1, M] where M is the maximum element of the array. Therefore, the idea is to iterate over then range [1, M] and if any element in the range is a factor of an array element then print the current element as one of the resultant GCD. Follow the steps below to solve the problem:
- Store all the array elements in the HashSet, say s.
- Iterate over the range [1, M] using the variable i and perform the following steps:
- Iterate through all the multiples of i, if there exists any multiple which is present in the HashSet, then print the current element i as one of the possible GCD.
Below is an implementation of the above approach:
// C++ program for the above approach
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the different GCDs of // the subsequence of the given array void findGCDsSubsequence(vector< int > arr) { // Stores all the possible GCDs vector< int > ans; // Stores all array element in set set< int > s; for ( int i : arr) s.insert(i); int M = *max_element(arr.begin(), arr.end()); // Iterate over the range [1, M] for ( int i = 1; i <= M; i++) { int gcd = 0; // Check if i can be the GCD of // any subsequence for ( int j = i; j < M + 1; j += i) { if (s.find(j) != s.end()) gcd = __gcd(gcd, j); } if (gcd == i) ans.push_back(i); } for ( int i = 0; i < ans.size(); i++) cout << ans[i] << " " ; } // Driver Code int main() { int N = 7; vector< int > arr = { 3, 4, 8 }; // Function Call findGCDsSubsequence(arr); return 0; } // This code is contributed by parthagarwal1962000 |
Java
// Java program for the above approach import java.util.*; public class GFG { static int gcd1( int a, int b) { return b == 0 ? a : gcd1(b, a % b); } // Function to find the different GCDs of // the subsequence of the given array static void findGCDsSubsequence(ArrayList<Integer> arr) { // Stores all the possible GCDs ArrayList<Integer> ans = new ArrayList<Integer>(); // Stores all array element in set HashSet<Integer> s = new HashSet<Integer>(); for ( int i : arr) s.add(i); int M = Integer.MIN_VALUE; for ( int i : arr) { if (i > M) M = i; } // Iterate over the range [1, M] for ( int i = 1 ; i <= M; i++) { int gcd = 0 ; // Check if i can be the GCD of // any subsequence for ( int j = i; j < M + 1 ; j += i) { if (s.contains(j)) gcd = gcd1(gcd, j); } if (gcd == i) ans.add(i); } for ( int i = 0 ; i < ans.size(); i++) System.out.print(ans.get(i) + " " ); } // Driver Code public static void main(String args[]) { ArrayList<Integer> arr = new ArrayList<Integer>(); arr.add( 3 ); arr.add( 4 ); arr.add( 8 ); // Function Call findGCDsSubsequence(arr); } } // This code is contributed by SoumikMondal |
Python3
# Python3 program for the above approach import math # Function to find the different GCDs of # the subsequence of the given array def findGCDsSubsequence(nums): # Stores all the possible GCDs Ans = [] # Stores all array element in set s = set (nums) # Find the maximum array element M = max (nums) # Iterate over the range [1, M] for i in range ( 1 , M + 1 ): # Stores the GCD of subsequence gcd = 0 # Check if i can be the GCD of # any subsequence for j in range (i, M + 1 , i): if j in s: gcd = math.gcd(gcd, j) # Store the value i in Ans[] # if it can be the GCD if gcd = = i: Ans + = [i] # Print all possible GCDs stored print ( * Ans) # Driver Code N = 7 arr = [ 3 , 4 , 8 ] # Function Call findGCDsSubsequence(arr) |
C#
// C# program for the above approach using System; using System.Collections.Generic; class GFG{ static int gcd1( int a, int b) { return b == 0 ? a : gcd1(b, a % b); } // Function to find the different GCDs of // the subsequence of the given array static void findGCDsSubsequence(List< int > arr) { // Stores all the possible GCDs List< int > ans = new List< int >(); // Stores all array element in set HashSet< int > s = new HashSet< int >(); foreach ( int i in arr) s.Add(i); int M = Int32.MinValue; foreach ( int i in arr) { if (i > M) M = i; } // Iterate over the range [1, M] for ( int i = 1; i <= M; i++) { int gcd = 0; // Check if i can be the GCD of // any subsequence for ( int j = i; j < M + 1; j += i) { if (s.Contains(j)) gcd = gcd1(gcd, j); } if (gcd == i) ans.Add(i); } for ( int i = 0; i < ans.Count; i++) Console.Write(ans[i] + " " ); } // Driver Code public static void Main() { List< int > arr = new List< int >(){ 3, 4, 8 }; // Function Call findGCDsSubsequence(arr); } } // This code is contributed by SURENDRA_GANGWAR |
Javascript
<script> // JavaScript program for the above approach function __gcd(a, b) { if (b == 0) return a; return __gcd(b, a % b); } // Function to find the different GCDs of // the subsequence of the given array function findGCDsSubsequence(arr) { // Stores all the possible GCDs let ans = []; // Stores all array element in set let s = new Set(); for (let i of arr) s.add(i); let M = [...arr].sort((a, b) => b - a)[0] // Iterate over the range [1, M] for (let i = 1; i <= M; i++) { let gcd = 0; // Check if i can be the GCD of // any subsequence for (let j = i; j < M + 1; j += i) { if (s.has(j)) gcd = __gcd(gcd, j); } if (gcd == i) ans.push(i); } for (let i = 0; i < ans.length; i++) document.write(ans[i] + " " ); } // Driver Code let N = 7; let arr = [3, 4, 8]; // Function Call findGCDsSubsequence(arr); // This code is contributed by gfgking </script> |
1 3 4 8
Time Complexity: O(M*log M), where M is the maximum element of the array.
Auxiliary Space: O(N)