# GCD of more than two (or array) numbers

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)**

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