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Largest number dividing maximum number of elements in the array

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++ implementation of the approach
#include <bits/stdc++.h>
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 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 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# 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




<script>
// javascript implementation of the approach   
// Function to return the largest number
    // that divides the maximum elements
    // from the given array
    function findLargest(arr , n) {
 
        // Finding gcd of all the numbers
        // in the array
        var gcd = 0;
        for (i = 0; i < n; i++)
            gcd = __gcd(arr[i], gcd);
        return gcd;
    }
 
    function __gcd(a , b) {
        return b == 0 ? a : __gcd(b, a % b);
    }
 
    // Driver code
     
        var arr = [ 3, 6, 9 ];
        var n = arr.length;
 
        document.write(findLargest(arr, n));
 
// This code contributed by umadevi9616
</script>

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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