Count of integers that divide all the elements of the given array
Given an array arr[] of N elements. The task is to find the count of positive integers that divide all the array elements.
Examples:
Input: arr[] = {2, 8, 10, 6}
Output: 2
1 and 2 are the only integers that divide
all the elements of the given array.Input: arr[] = {6, 12, 18, 12, 6}
Output: 4
Approach: We know that the maximum integer that will divide all the array elements will be the gcd of the array and all the other integers that will divide all the elements of the array will have to be the factors of this gcd. Hence, the count of valid integers will be equal to the count of factors of the gcd of all the array elements.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count of // the required integers int getCount( int a[], int n) { // To store the gcd of the array elements int gcd = 0; for ( int i = 0; i < n; i++) gcd = __gcd(gcd, a[i]); // To store the count of factors // of the found gcd int cnt = 0; for ( int i = 1; i * i <= gcd; i++) { if (gcd % i == 0) { // If g is a perfect square if (i * i == gcd) cnt++; // Factors appear in pairs else cnt += 2; } } return cnt; } // Driver code int main() { int a[] = { 4, 16, 1024, 48 }; int n = sizeof (a) / sizeof (a[0]); cout << getCount(a, n); return 0; } |
Java
// Java implementation of the approach class GFG { // Recursive function to return gcd static int calgcd( int a, int b) { if (b == 0 ) return a; return calgcd(b, a % b); } // Function to return the count of // the required integers static int getCount( int [] a, int n) { // To store the gcd of the array elements int gcd = 0 ; for ( int i = 0 ; i < n; i++) gcd = calgcd(gcd, a[i]); // To store the count of factors // of the found gcd int cnt = 0 ; for ( int i = 1 ; i * i <= gcd; i++) { if (gcd % i == 0 ) { // If g is a perfect square if (i * i == gcd) cnt++; // Factors appear in pairs else cnt += 2 ; } } return cnt; } // Driver code public static void main (String[] args) { int [] a = { 4 , 16 , 1024 , 48 }; int n = a.length; System.out.println(getCount(a, n)); } } // This code is contributed by ihritik |
Python3
# Python3 implementation of the approach # Function to return the count of # the required integers from math import gcd as __gcd def getCount(a, n): # To store the gcd of the array elements gcd = 0 for i in range (n): gcd = __gcd(gcd, a[i]) # To store the count of factors # of the found gcd cnt = 0 for i in range ( 1 , gcd + 1 ): if i * i > gcd: break if (gcd % i = = 0 ): # If g is a perfect square if (i * i = = gcd): cnt + = 1 # Factors appear in pairs else : cnt + = 2 return cnt # Driver code a = [ 4 , 16 , 1024 , 48 ] n = len (a) print (getCount(a, n)) # This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System; class GFG { // Recursive function to return gcd static int calgcd( int a, int b) { if (b == 0) return a; return calgcd(b, a % b); } // Function to return the count of // the required integers static int getCount( int [] a, int n) { // To store the gcd of the array elements int gcd = 0; for ( int i = 0; i < n; i++) gcd = calgcd(gcd, a[i]); // To store the count of factors // of the found gcd int cnt = 0; for ( int i = 1; i * i <= gcd; i++) { if (gcd % i == 0) { // If g is a perfect square if (i * i == gcd) cnt++; // Factors appear in pairs else cnt += 2; } } return cnt; } // Driver code public static void Main () { int [] a = { 4, 16, 1024, 48 }; int n = a.Length; Console.WriteLine(getCount(a, n)); } } // This code is contributed by ihritik |
Javascript
<script> // Javascript implementation of the approach function calgcd(a, b) { if (b == 0) return a; return calgcd(b, a % b); } // Function to return the count of // the required integers function getCount(a, n) { // To store the gcd of the array elements let gcd = 0; for (let i = 0; i < n; i++) gcd = calgcd(gcd, a[i]); // To store the count of factors // of the found gcd let cnt = 0; for (let i = 1; i * i <= gcd; i++) { if (gcd % i == 0) { // If g is a perfect square if (i * i == gcd) cnt++; // Factors appear in pairs else cnt += 2; } } return cnt; } // Driver code let a = [ 4, 16, 1024, 48 ]; let n = a.length; document.write(getCount(a, n)); </script> |
Output:
3
Time Complexity: O(N*log(max_element))
Auxiliary Space: O(1)
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