Find integers that divides maximum number of elements of the array

Given an array arr[] of integers, the task is to find the element (other than 1) which is the factor of the maximum number of elements in the array. If multiple such factors exist, print all the factors in ascending order.

Examples:

Input: arr[] = {10, 20} 
Output: 2 5 10 
The factors of 10 are 1, 2, 5, 10. 
The factors of 20 are 1, 2, 4, 5, 10, 20. 
The factors other than 1, which occur the most times (twice) are 2, 5, 10.

Input: arr[] = {120, 15, 24, 63, 18} 
Output: 3 

Approach:

• Initialize two lists, one to store the rank (the number of elements that the integer is a factor of) of the factor and another to store the factor.
• Start from 2 till the maximum element of the array.
• Count the number of elements in the array. The current integer is a factor of.
• Add the count to the rank list and the integers to the factor list.
• Find the integer with the maximum rank.
• Print all the elements with the same rank.

Below is the implementation of the above approach: 

C++

// CPP implementation of the approach
#include<bits/stdc++.h>
using namespace std;

// Function to print the integers that divide
// the maximum number of elements from the array
void maximumFactor(vector<int>arr)
{
    // Initialize two lists
    // to store rank and factors
    int n = arr.size();
    vector<int> rank;
    vector<int> factors;
    int max = *max_element(arr.begin(), arr.end());

    // Start from 2 till the maximum element in arr
    for (int i = 2; i <= max; i++)
    {
        // Initialize a variable
        // to count the number of elements
        // it is a factor of
        int count = 0;
        for (int j = 0; j < n; j++)
        {
            if (arr[j] % i == 0)
                count+= 1;
            rank.push_back(count);
            factors.push_back(i);
        }

    }



    // Maximum rank in the rank list
    int m = *max_element(rank.begin(),rank.end());
    for (int i = 0; i < rank.size(); i++)
    {
        // Print all the elements with rank m
        if (rank[i] == m)
            cout << factors[i] <<" ";
    }

}

// Driver code
int main()
{
    vector<int>arr = {120, 15, 24, 63, 18};
    maximumFactor(arr);
}

// This code is contributed by
// Surendra_Gangwar

Java

// Java implementation of the approach
import java.util.*;
class GFG
{

// Function to print the integers that
// divide the maximum number of
// elements from the array
static void maximumFactor(int []arr)
{

    // Initialize two lists to store
    // rank and factors
    int[] rank = new int[Arrays.stream(arr).max().getAsInt() + 1];
    int[] factors = new int[Arrays.stream(arr).max().getAsInt() + 1];
    int g = 0;

    // Start from 2 till the maximum
    // element in arr
    for (int i = 2;
             i <= Arrays.stream(arr).max().getAsInt(); i++)
    {
        // Initialize a variable to count
        // the number of elements it is a
        // factor of
        int count = 0;
        for (int j = 0; j < arr.length; j++)
            if (arr[j] % i == 0)
                count += 1;

        rank[g] = count;
        factors[g] = i;
        g++;
    }

    // Maximum rank in the rank list
    int m = Arrays.stream(rank).max().getAsInt();
    for (int i = 0; i < rank.length; i++)
    {
        // Print all the elements with rank m
        if (rank[i] == m)
            System.out.print(factors[i] + " ");
    }
}

// Driver code
public static void main (String[] args)
{
    int []arr = {120, 15, 24, 63, 18};
    maximumFactor(arr);
}
}

// This code is contributed by
// chandan_jnu

Python

# Python3 implementation of the approach

# Function to print the integers that divide
# the maximum number of elements from the array
def maximumFactor(arr):

    # Initialize two lists
    # to store rank and factors
    rank, factors = [], []

    # Start from 2 till the maximum element in arr
    for i in range(2, max(arr)+1):

        # Initialize a variable
        # to count the number of elements
        # it is a factor of
        count = 0
        for j in arr:
            if j % i == 0:count+= 1
        rank.append(count)
        factors.append(i)

    # Maximum rank in the rank list
    m = max(rank)
    for i in range(len(rank)):

        # Print all the elements with rank m
        if rank[i]== m:
            print(factors[i], end =" ")

# Driver code
arr = [120, 15, 24, 63, 18]

maximumFactor(arr)

C#

// C# implementation of the approach
using System;
using System.Collections;
using System.Linq;

class GFG
{

// Function to print the integers that
// divide the maximum number of
// elements from the array
static void maximumFactor(int []arr)
{

    // Initialize two lists to store
    // rank and factors
    int[] rank = new int[arr.Max() + 1];
    int[] factors = new int[arr.Max() + 1];
    int g = 0;

    // Start from 2 till the maximum
    // element in arr
    for (int i = 2; i <= arr.Max(); i++)
    {
        // Initialize a variable to count
        // the number of elements it is a
        // factor of
        int count = 0 ;
        for (int j = 0; j < arr.Length; j++)
            if (arr[j] % i == 0)
                count += 1;

        rank[g]=count;
        factors[g]=i;
        g++;
    }

    // Maximum rank in the rank list
    int m = rank.Max();
    for (int i = 0; i < rank.Length; i++)
    {
        // Print all the elements with rank m
        if ((int)rank[i] == m)
            Console.Write(factors[i]+" ");
    }
}

// Driver code
static void Main()
{

int []arr = {120, 15, 24, 63, 18};
maximumFactor(arr);
}
}

// This code is contributed by chandan_jnu

PHP

<?php
// PHP implementation of the approach

// Function to print the integers that
// divide the maximum number of
// elements from the array
function maximumFactor($arr)
{

    // Initialize two lists to store
    // rank and factors
    $rank = array();
    $factors = array();

    // Start from 2 till the maximum
    // element in arr
    for ($i = 2; $i <= max($arr); $i++)
    {
        // Initialize a variable to count
        // the number of elements it is a
        // factor of
        $count = 0 ;
        for ($j = 0; $j < sizeof($arr); $j++)
            if ($arr[$j] % $i == 0)
                $count += 1;

        array_push($rank, $count);
        array_push($factors, $i);
    }

    // Maximum rank in the rank list
    $m = max($rank);
    for ($i = 0; $i < sizeof($rank); $i++)
    {
        // Print all the elements with rank m
        if ($rank[$i] == $m)
            echo $factors[$i], " ";
    }
}

// Driver code
$arr = array(120, 15, 24, 63, 18);

maximumFactor($arr)

// This code is contributed by Ryuga
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to print the integers that divide
// the maximum number of elements from the array
function maximumFactor(arr)
{
    // Initialize two lists
    // to store rank and factors
    var n = arr.length;
    var rank = [];
    var factors = [];
    var max = arr.reduce((a,b)=> Math.max(a,b));

    // Start from 2 till the maximum element in arr
    for (var i = 2; i <= max; i++)
    {
        // Initialize a variable
        // to count the number of elements
        // it is a factor of
        var count = 0;
        for (var j = 0; j < n; j++)
        {
            if (arr[j] % i == 0)
                count+= 1;
            rank.push(count);
            factors.push(i);
        }

    }



    // Maximum rank in the rank list
    var m = rank.reduce((a,b)=>Math.max(a,b));
    for (var i = 0; i < rank.length; i++)
    {
        // Print all the elements with rank m
        if (rank[i] == m)
            document.write( factors[i] + " ");
    }

}

// Driver code
var arr = [120, 15, 24, 63, 18];
maximumFactor(arr);


</script>

Output

3 

Complexity Analysis:

• Time Complexity: O(n * max(arr)), where max(arr) is the largest element of the array arr.
• Auxiliary Space: O(n * max(arr))