Given an array of numbers, find GCD of the array elements. In a previous post we find GCD of two number.
Examples:
Input : arr[] = {1, 2, 3}
Output : 1
Input : arr[] = {2, 4, 6, 8}
Output : 2
The GCD of three or more numbers equals the product of the prime factors common to all the numbers, but it can also be calculated by repeatedly taking the GCDs of pairs of numbers.
gcd(a, b, c) = gcd(a, gcd(b, c))
= gcd(gcd(a, b), c)
= gcd(gcd(a, c), b)
For an array of elements, we do the following. We will also check for the result if the result at any step becomes 1 we will just return the 1 as gcd(1,x)=1.
result = arr[0] For i = 1 to n-1 result = GCD(result, arr[i])
Below is the implementation of the above idea.
C++
// C++ program to find GCD of two or // more numbers #include <bits/stdc++.h> using namespace std; // Function to return gcd of a and b int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Function to find gcd of array of // numbers int findGCD(int arr[], int n) { int result = arr[0]; for (int i = 1; i < n; i++) { result = gcd(arr[i], result); if(result == 1) { return 1; } } return result; } // Driver code int main() { int arr[] = { 2, 4, 6, 8, 16 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findGCD(arr, n) << endl; return 0; } |
Java
// Java program to find GCD of two or // more numbers public class GCD { // Function to return gcd of a and b static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Function to find gcd of array of // numbers static int findGCD(int arr[], int n) { int result = arr[0]; for (int element: arr){ result = gcd(result, element); if(result == 1) { return 1; } } return result; } public static void main(String[] args) { int arr[] = { 2, 4, 6, 8, 16 }; int n = arr.length; System.out.println(findGCD(arr, n)); } } // This code is contributed by Saket Kumar |
Python
# GCD of more than two (or array) numbers # Function implements the Euclidean # algorithm to find H.C.F. of two number def find_gcd(x, y): while(y): x, y = y, x % y return x # Driver Code l = [2, 4, 6, 8, 16] num1 = l[0] num2 = l[1] gcd = find_gcd(num1, num2) for i in range(2, len(l)): gcd = find_gcd(gcd, l[i]) print(gcd) # Code contributed by Mohit Gupta_OMG |
C#
// C# program to find GCD of // two or more numbers using System; public class GCD { // Function to return gcd of a and b static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Function to find gcd of // array of numbers static int findGCD(int[] arr, int n) { int result = arr[0]; for (int i = 1; i < n; i++){ result = gcd(arr[i], result); if(result == 1) { return 1; } } return result; } // Driver Code public static void Main() { int[] arr = { 2, 4, 6, 8, 16 }; int n = arr.Length; Console.Write(findGCD(arr, n)); } } // This code is contributed by nitin mittal |
PHP
<?php // PHP program to find GCD of two or // more numbers // Function to return gcd of a and b function gcd( $a, $b) { if ($a == 0) return $b; return gcd($b % $a, $a); } // Function to find gcd of array of // numbers function findGCD($arr, $n) { $result = $arr[0]; for ($i = 1; $i < $n; $i++){ $result = gcd($arr[$i], $result); if($result == 1) { return 1; } } return $result; } // Driver code $arr = array( 2, 4, 6, 8, 16 ); $n = sizeof($arr); echo (findGCD($arr, $n)); // This code is contributed by // Prasad Kshirsagar. ?> |
Javascript
<script> // Javascript program to find GCD of two or // more numbers // Function to return gcd of a and b function gcd(a, b) { if (a == 0) return b; return gcd(b % a, a); } // Function to find gcd of array of // numbers function findGCD(arr, n) { let result = arr[0]; for (let i = 1; i < n; i++) { result = gcd(arr[i], result); if(result == 1) { return 1; } } return result; } // Driver code let arr = [ 2, 4, 6, 8, 16 ]; let n = arr.length; document.write(findGCD(arr, n) + "<br>"); // This is code is contributed by Mayank Tyagi </script> |
Output
2
Time Complexity: O(N * log(N)), where N is the largest element of the array
Auxiliary Space: O(1)
Recursive Method: Implementation of Algorithm recursively :
C++
#include <bits/stdc++.h> using namespace std; // recursive implementation int GcdOfArray(vector<int>& arr, int idx) { if (idx == arr.size() - 1) { return arr[idx]; } int a = arr[idx]; int b = GcdOfArray(arr, idx + 1); return __gcd( a, b); // __gcd(a,b) is inbuilt library function } int main() { vector<int> arr = { 1, 2, 3 }; cout << GcdOfArray(arr, 0) << "\n"; arr = { 2, 4, 6, 8 }; cout << GcdOfArray(arr, 0) << "\n"; return 0; } |
Java
import java.util.*; class GFG { // Recursive function to return gcd of a and b static int __gcd(int a, int b) { return b == 0? a:__gcd(b, a % b); } // recursive implementation static int GcdOfArray(int[] arr, int idx) { if (idx == arr.length - 1) { return arr[idx]; } int a = arr[idx]; int b = GcdOfArray(arr, idx + 1); return __gcd( a, b); // __gcd(a,b) is inbuilt library function } public static void main(String[] args) { int[] arr = { 1, 2, 3 }; System.out.print(GcdOfArray(arr, 0)+ "\n"); int[] arr1 = { 2, 4, 6, 8 }; System.out.print(GcdOfArray(arr1, 0)+ "\n"); } } // This code is contributed by gauravrajput1 |
Python3
# To find GCD of an array by recursive approach import math # Recursive Implementation def GcdOfArray(arr, idx): if idx == len(arr)-1: return arr[idx] a = arr[idx] b = GcdOfArray(arr,idx+1) return math.gcd(a, b) # Driver Code arr = [1, 2, 3] print(GcdOfArray(arr,0)) arr = [2, 4, 6, 8] print(GcdOfArray(arr,0)) # Code contributed by Gautam goel (gautamgoel962) |
C#
// To find GCD of an array by recursive approach using System; public class GCD { // Function to return gcd of a and b static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Recursive Implementation static int GcdOfArray(int[] arr, int idx) { if(idx==arr.Length-1) { return arr[idx]; } int a = arr[idx]; int b = GcdOfArray(arr, idx + 1); return gcd(a,b); } // Driver Code public static void Main() { int[] arr = { 1,2,3 }; Console.Write(GcdOfArray(arr, 0)+"\n"); int[] arr1 = { 2,4,6,8 }; Console.Write(GcdOfArray(arr1, 0)); } } // This code is contributed by adityapatil12 |
Javascript
// JavaScript code to implement this approach // function to calculate gcd function gcd(a, b) { if (!b) return a; return gcd(b, a % b); } // recursive implementation function GcdOfArray(arr, idx) { if (idx == arr.length - 1) { return arr[idx]; } var a = arr[idx]; var b = GcdOfArray(arr, idx + 1); return gcd(a, b); } // Driver Code var arr = [ 1, 2, 3 ]; console.log(GcdOfArray(arr, 0)); arr = [ 2, 4, 6, 8 ]; console.log(GcdOfArray(arr, 0)); // This code is contributed by phasing17 |
Output
1 2
Time Complexity: O(N * log(N)), where N is the largest element of the array
Auxiliary Space: O(N)
Iterative version of the above solution
C++
#include <bits/stdc++.h> using namespace std; // iterative implementation int getGCD(int a, int b) { while (a > 0 && b > 0) { if (a > b) { a = a % b; } else { b = b % a; } } if (a == 0) { return b; } return a; } int GcdOfArray(vector<int>& arr) { int gcd = 0; for (int i = 0; i < arr.size(); i++) { gcd = getGCD(gcd, arr[i]); } return gcd; } int main() { vector<int> arr = { 1, 2, 3 }; cout << GcdOfArray(arr) << "\n"; arr = { 2, 4, 6, 8 }; cout << GcdOfArray(arr) << "\n"; return 0; } |
Java
import java.util.*; class GFG { // iterative implementationto return gcd of a and b static int getGCD(int a, int b) { while (a > 0 && b > 0) { if (a > b) { a = a % b; } else { b = b % a; } } if (a == 0) { return b; } return a; } // recursive implementation static int GcdOfArray(int[] arr) { int gcd = 0; for (int i = 0; i < arr.length; i++) { gcd = getGCD(gcd, arr[i]); } return gcd; } public static void main(String[] args) { int[] arr = { 1, 2, 3 }; System.out.print(GcdOfArray(arr)+ "\n"); int[] arr1 = { 2, 4, 6, 8 }; System.out.print(GcdOfArray(arr1)+ "\n"); } } |
Python
# Python implementation of above approach # iterative implementation def getGCD(a, b): while(a > 0 and b > 0): if(a > b): a = a % b else: b = b % a if(a == 0): return b return a def GcdOfArray(arr): gcd = 0 for i in range(len(arr)): gcd = getGCD(gcd, arr[i]) return gcd # driver code to test above functions arr = [1,2,3] print(GcdOfArray(arr)) arr = [2,4,6,8] print(GcdOfArray(arr)) # THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002) |
C#
// To find GCD of an array by recursive approach using System; public class GCD { // Function to return gcd of a and b static int getGCD(int a, int b) { while (a > 0 && b > 0) { if (a > b) { a = a % b; } else { b = b % a; } } if (a == 0) { return b; } return a; } // Recursive Implementation static int GcdOfArray(int[] arr) { int gcd = 0; for (int i = 0; i < arr.Length; i++) { gcd = getGCD(gcd, arr[i]); } return gcd; } // Driver Code public static void Main() { int[] arr = { 1, 2, 3 }; Console.Write(GcdOfArray(arr) + "\n"); int[] arr1 = { 2, 4, 6, 8 }; Console.Write(GcdOfArray(arr1)); } } |
Javascript
// JavaScript code to implement this approach // function to calculate gcd function getGCD( a, b) { while (a > 0 && b > 0) { if (a > b) { a = a % b; } else { b = b % a; } } if (a == 0) { return b; } return a; } // iterative implementation function GcdOfArray(arr) { let gcd = 0; for (int i = 0; i < arr.size(); i++) { gcd = getGCD(gcd, arr[i]); } return gcd; } // Driver Code var arr = [ 1, 2, 3 ]; console.log(GcdOfArray(arr)); arr = [ 2, 4, 6, 8 ]; console.log(GcdOfArray(arr)); // This code is contributed by phasing17 |
Output
1 2
Time Complexity: O(N * log(X)), where X is the largest element of the array
Auxiliary Space: O(1)
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