# Largest number dividing maximum number of elements in the array

• Difficulty Level : Basic
• Last Updated : 19 Aug, 2021

Given an array arr[] of length N, the task is to find the largest number dividing the maximum number of elements from the array.

Examples:

Input: arr[] = {2, 12, 6}
Output:
1 and 2 are the only integers which divide the
maximum number of elements from the array
(i.e. all the elements) and 2 is
the maximum among them.
Input: arr[] = {1, 7, 9}
Output:

Approach: A straightforward approach for solving this problem will be taking the GCD of all the elements. Why this approach works? 1 is the number that divides all the elements of the array. Now, any other number greater than 1 will either divide all the elements of the array (in this case, the number itself is the answer) or it will divide a subset of the array i.e. 1 is the answer here as it divides more elements from the array. So, the most straightforward way for doing this will be to take the GCD of all the elements of the array.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the largest number``// that divides the maximum elements``// from the given array``int` `findLargest(``int``* arr, ``int` `n)``{` `    ``// Finding gcd of all the numbers``    ``// in the array``    ``int` `gcd = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``gcd = __gcd(arr[i], gcd);``    ``return` `gcd;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 3, 6, 9 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);` `    ``cout << findLargest(arr, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG {` `    ``// Function to return the largest number``    ``// that divides the maximum elements``    ``// from the given array``    ``static` `int` `findLargest(``int``[] arr, ``int` `n)``    ``{` `        ``// Finding gcd of all the numbers``        ``// in the array``        ``int` `gcd = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``gcd = __gcd(arr[i], gcd);``        ``return` `gcd;``    ``}` `    ``static` `int` `__gcd(``int` `a, ``int` `b)``    ``{``        ``return` `b == ``0` `? a : __gcd(b, a % b);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``3``, ``6``, ``9` `};``        ``int` `n = arr.length;` `        ``System.out.print(findLargest(arr, n));``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 implementation of the approach``from` `math ``import` `gcd as __gcd` `# Function to return the largest number``# that divides the maximum elements``# from the given array``def` `findLargest(arr, n):` `    ``# Finding gcd of all the numbers``    ``# in the array``    ``gcd ``=` `0``    ``for` `i ``in` `range``(n):``        ``gcd ``=` `__gcd(arr[i], gcd)``    ``return` `gcd` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``3``, ``6``, ``9``]``    ``n ``=` `len``(arr)` `    ``print``(findLargest(arr, n))` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG {` `    ``// Function to return the largest number``    ``// that divides the maximum elements``    ``// from the given array``    ``static` `int` `findLargest(``int``[] arr, ``int` `n)``    ``{` `        ``// Finding gcd of all the numbers``        ``// in the array``        ``int` `gcd = 0;``        ``for` `(``int` `i = 0; i < n; i++)``            ``gcd = __gcd(arr[i], gcd);``        ``return` `gcd;``    ``}` `    ``static` `int` `__gcd(``int` `a, ``int` `b)``    ``{``        ``return` `b == 0 ? a : __gcd(b, a % b);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] arr = { 3, 6, 9 };``        ``int` `n = arr.Length;` `        ``Console.Write(findLargest(arr, n));``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

`3`

Time Complexity: O(N * log(MAX)), where N is the size of the array and MAX is the maximum element of the array.
Auxiliary Space: O(log(MAX))

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